{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"checkbox","correct":[["1","2"]],"list":[{"point":5,"img":"","ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=a{{x}^{2}}.$ Bi\u1ebft \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i qua $A(-1;2)$<br\/><b>C\u00e2u a:<\/b> Tr\u00ean \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1, \u0111i\u1ec3m c\u00f3 tung \u0111\u1ed9 b\u1eb1ng $4$ l\u00e0:<\/span>","hint":"","column":2,"number_true":2,"select":["A. $\\left( \\sqrt{2};4 \\right)$","B. $\\left( -\\sqrt{2};4 \\right)$","C. $\\left( -2;4 \\right)$","D. $\\left( 2;4 \\right)$"],"explain":"<span class='basic_left'>\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i qua $A(-1;2)$ n\u00ean thay $x=-1$ v\u00e0 $y=2$ v\u00e0o h\u00e0m s\u1ed1 $y=ax^2,$ ta c\u00f3: <br\/>$2=a.{{\\left( -1 \\right)}^{2}}\\Leftrightarrow a=2$ <br\/>Khi \u0111\u00f3, ta c\u00f3 h\u00e0m s\u1ed1: $y=2x^2$<br\/>C\u00e1c \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb c\u00f3 tung \u0111\u1ed9 b\u1eb1ng $4$ n\u00ean $2{{x}^{2}}=4\\Leftrightarrow {{x}^{2}}=2\\Leftrightarrow x=\\pm \\sqrt{2}$ <br\/>V\u1eady c\u00e1c \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb c\u00f3 tung \u0111\u1ed9 b\u1eb1ng $4$ l\u00e0 $\\left( \\sqrt{2};4 \\right)$ v\u00e0 $\\left( -\\sqrt{2};4 \\right)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 A v\u00e0 B.<\/span><\/span> "}]}],"id_ques":1041},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"checkbox","correct":[["1","2","4"]],"list":[{"point":5,"img":"","ques":"<span class='basic_left'>Cho h\u00e0m s\u1ed1 $y=a{{x}^{2}}.$ Bi\u1ebft \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i qua $A(-1;2)$<br\/><b>C\u00e2u b:<\/b> C\u00e1c \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb v\u00e0 c\u00e1ch \u0111\u1ec1u hai tr\u1ee5c t\u1ecda \u0111\u1ed9 l\u00e0:<\/span>","hint":"C\u00e1c \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u hai tr\u1ee5c t\u1ecda \u0111\u1ed9 th\u00ec n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $y=x$ v\u00e0 $y=-x$","column":2,"number_true":2,"select":["A. $O(0;0)$","B. $\\left( \\dfrac{1}{2};\\dfrac{1}{2} \\right)$","C. $\\left( \\dfrac{1}{2};-\\dfrac{1}{2} \\right)$","D. $\\left( -\\dfrac{1}{2};\\dfrac{1}{2} \\right)$"],"explain":"<span class='basic_left'>T\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u hai tr\u1ee5c t\u1ecda \u0111\u1ed9 l\u00e0 c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $y=x$ v\u00e0 $y=-x$<br\/>Theo <b>c\u00e2u a <\/b>, \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 \u0111i qua $A(-1;2)$ l\u00e0 parabol $y=2x^2$ (P)<br\/>Khi \u0111\u00f3 t\u1ecda \u0111\u1ed9 c\u00e1c \u0111i\u1ec3m tr\u00ean \u0111\u1ed3 th\u1ecb (P) v\u00e0 c\u00e1ch \u0111\u1ec1u hai tr\u1ee5c t\u1ecda \u0111\u1ed9 l\u00e0 nghi\u1ec7m c\u1ee7a c\u00e1c h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\left\\{ \\begin{aligned} & y=2{{x}^{2}} \\\\ & y=x \\\\ \\end{aligned} \\right.$ (I) v\u00e0 $\\left\\{ \\begin{aligned} & y=2{{x}^{2}} \\\\ & y=-x \\\\ \\end{aligned} \\right.$ (II)<br\/>Gi\u1ea3i h\u1ec7 (I)<br\/>(I)$\\Leftrightarrow \\left\\{ \\begin{aligned} & 2{{x}^{2}}=x \\\\ & y=x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2{{x}^{2}}-x=0 \\\\ & y=x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\left( 2x-1 \\right)=0 \\\\ & y=x \\\\ \\end{aligned} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\left[ \\begin{aligned} & x=0 \\\\ & x=\\dfrac{1}{2} \\\\ \\end{aligned} \\right. \\\\ & y=x \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=y=0 \\\\ & x=y=\\dfrac{1}{2} \\\\ \\end{aligned} \\right.$<br\/> T\u01b0\u01a1ng t\u1ef1 gi\u1ea3i h\u1ec7 (II), ta \u0111\u01b0\u1ee3c: $x=y=0$ v\u00e0 $x=-\\dfrac{1}{2};y=\\dfrac{1}{2}$<br\/>V\u1eady c\u00e1c \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u hai tr\u1ee5c t\u1ecda \u0111\u1ed9 l\u00e0 $O(0;0),$ $A\\left( \\dfrac{1}{2};\\dfrac{1}{2} \\right)$ v\u00e0 $B\\left( -\\dfrac{1}{2};\\dfrac{1}{2} \\right)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 A, B v\u00e0 D.<\/span><\/span> "}]}],"id_ques":1042},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["f","f","t","t"]],"list":[{"point":5,"image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["N\u1ebfu \u0111i\u1ec3m $M(a;b)$ thu\u1ed9c (P): $y=ax^2,$ $a$ kh\u00e1c $0$ th\u00ec \u0111i\u1ec3m $M\u2019(a;-b)$ c\u0169ng thu\u1ed9c (P)","\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 $\\Delta >0$","N\u1ebfu ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ ($a\\ne 0$ ) c\u00f3 $ac<0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t.","Hai ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0\\,\\,\\left( a\\ne 0 \\right)$ v\u00e0 $a{{x}^{2}}-bx+c=0\\,\\,\\left( a\\ne 0 \\right)$ c\u00f9ng c\u00f3 nghi\u1ec7m ho\u1eb7c c\u00f9ng v\u00f4 nghi\u1ec7m"],"hint":"","explain":["Sai v\u00ec \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 $y=a{{x}^{2}}\\left( a\\ne 0 \\right)$ \u0111i qua $O$ v\u00e0 nh\u1eadn $Oy$ l\u00e0m tr\u1ee5c \u0111\u1ed1i x\u1ee9ng n\u00ean \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $M(a;b)$ thu\u1ed9c (P) l\u00e0 \u0111i\u1ec3m $N(-a;b)$ thu\u1ed9c (P) ch\u1ee9 kh\u00f4ng ph\u1ea3i \u0111i\u1ec3m $M\u2019(a;-b)$","<br\/>Sai v\u00ec \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00f3 ph\u1ea3i l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 $\\Delta >0$ hay $\\left\\{ \\begin{align} & a\\ne 0 \\\\ & \\Delta >0 \\\\ \\end{align} \\right.$ ","<br\/>\u0110\u00fang v\u00ec v\u1edbi $ac<0$ th\u00ec $\\Delta ={{b}^{2}}-4ac>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t","\u0110\u00fang v\u00ec c\u1ea3 hai ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec1u c\u00f3 $\\Delta ={{b}^{2}}-4ac$ n\u00ean c\u1ea3 hai ph\u01b0\u01a1ng tr\u00ecnh c\u00f9ng c\u00f3 nghi\u1ec7m ho\u1eb7c c\u00f9ng v\u00f4 nghi\u1ec7m."]}]}],"id_ques":1043},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["-1"],["2"],["1"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Cho parabol (P): $y=\\dfrac{{{x}^{2}}}{4}$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=x+m$<br\/><b> C\u00e2u a: <\/b> \u0110\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng $y=x+m$ ti\u1ebfp x\u00fac v\u1edbi parabol th\u00ec $m=$_input_. T\u1ecda \u0111\u1ed9 ti\u1ebfp \u0111i\u1ec3m l\u00e0 (_input_;_input_) <\/span>","hint":"\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 (d) ti\u1ebfp x\u00fac v\u1edbi (P) l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u00f3 nghi\u1ec7m k\u00e9p","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (P) l\u00e0:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\dfrac{{{x}^{2}}}{4}=x+m\\,\\,\\,\\,\\left( * \\right) \\\\ & \\Leftrightarrow {{x}^{2}}-4x-4m=0 \\\\ & \\Delta '=4+4m \\\\ \\end{aligned}$ <br\/>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng (d) ti\u1ebfp x\u00fac v\u1edbi parabol (P) l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 nghi\u1ec7m k\u00e9p $\\Leftrightarrow \\Delta '=0\\Leftrightarrow 4+4m=0\\Leftrightarrow m=-1$ <br\/>V\u1edbi $m=-1$ th\u00ec \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=x-1$ ti\u1ebfp x\u00fac v\u1edbi parabol (P). Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (*) tr\u1edf th\u00e0nh:<br\/>$\\dfrac{{{x}^{2}}}{4}=x-1\\Leftrightarrow {{x}^{2}}-4x+4=0\\Leftrightarrow {{\\left( x-2 \\right)}^{2}}=0\\Leftrightarrow x=2$ <br\/>Thay $x=2$ v\u00e0o (d), ta c\u00f3: $y=2-1=1$<br\/>V\u1eady t\u1ecda \u0111\u1ed9 ti\u1ebfp \u0111i\u1ec3m l\u00e0 $(2;1)$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $-1;$ $2;$ $1.$<\/span><\/span>"}]}],"id_ques":1044},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho parabol (P):$y=\\dfrac{{{x}^{2}}}{4}$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=x+m$<br\/><b> C\u00e2u b: <\/b> T\u00ecm $m$ \u0111\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng (d) c\u1eaft parabol (P) t\u1ea1i 2 \u0111i\u1ec3m ph\u00e2n bi\u1ec7t n\u1eb1m b\u00ean ph\u1ea3i tr\u1ee5c tung<\/span>","select":["A. $m>0$ ","B. $m>-1$","C. $ -1 < m < 0 $","D. $m>0$ ho\u1eb7c $m<-1$ "],"hint":"(d) c\u1eaft (P) t\u1ea1i 2 \u0111i\u1ec3m ph\u00e2n bi\u1ec7t n\u1eb1m b\u00ean ph\u1ea3i tr\u1ee5c tung, n\u00ean hai \u0111i\u1ec3m \u0111\u1ec1u mang ho\u00e0nh \u0111\u1ed9 d\u01b0\u01a1ng. Khi \u0111\u00f3 b\u00e0i to\u00e1n tr\u1edf th\u00e0nh t\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t.","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (P) l\u00e0:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\dfrac{{{x}^{2}}}{4}=x+m \\\\ & \\Leftrightarrow {{x}^{2}}-4x-4m=0\\,\\,\\,\\,\\left( * \\right) \\\\ & \\Delta '=4+4m \\\\ \\end{aligned}$<br\/>$S=-\\dfrac{b}{a}=4;P=\\dfrac{c}{a}=-4m$ <br\/>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng (d) c\u1eaft parabol (P) t\u1ea1i 2 \u0111i\u1ec3m ph\u00e2n bi\u1ec7t n\u1eb1m b\u00ean ph\u1ea3i tr\u1ee5c tung l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t c\u00f9ng d\u01b0\u01a1ng<br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & \\Delta '>0 \\\\ & S>0 \\\\ & P>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 4+4m > 0 \\\\ & 4 > 0 \\\\ & -4m > 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m > -1 \\\\ & m < 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow -1 < m < 0 $ <br\/>V\u1eady $ -1 < m < 0 $ th\u00ec \u0111\u01b0\u1eddng th\u1eb3ng (d) c\u1eaft parabol (P) t\u1ea1i 2 \u0111i\u1ec3m ph\u00e2n bi\u1ec7t n\u1eb1m b\u00ean ph\u1ea3i tr\u1ee5c tung.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C. <\/span><\/span>","column":2}]}],"id_ques":1045},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["2","4"]],"list":[{"point":10,"img":"","ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}-mx+\\left( m-2 \\right)=0.$ T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $x_1$ v\u00e0 $x_2$ th\u1ecfa m\u00e3n $x_{1}^{2}+x_{2}^{2}=12$ ","hint":"","column":2,"number_true":2,"select":["A. $m=2$","B. $m=4$","C. $m=-1$","D. $m=-2$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m<br\/>B\u01b0\u1edbc 2: Vi\u1ebft h\u1ec7 th\u1ee9c Vi-\u00e9t. \u0110\u01b0a h\u1ec7 th\u1ee9c \u0111\u00e3 cho v\u1ec1 h\u1ec7 th\u1ee9c ch\u1ec9 ch\u1ee9a $x_1+x_2$ v\u00e0 $x_1.x_2$ r\u1ed3i \u00e1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t tr\u00ean \u0111\u1ec3 t\u00ecm $m.$<br\/>B\u01b0\u1edbc 3: Ki\u1ec3m tra $m$ c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n kh\u00f4ng r\u1ed3i k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: $\\Delta ={{m}^{2}}-4\\left( m-2 \\right)={{m}^{2}}-4m+8$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $x_1$ v\u00e0 $x_2$ $\\Leftrightarrow \\Delta \\ge 0\\Leftrightarrow {{m}^{2}}-4m+8\\ge 0$(*)<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=m \\\\ & {{x}_{1}}.{{x}_{2}}=m-2 \\\\ \\end{align} \\right.$ <br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,x_{1}^{2}+x_{2}^{2}=12 \\\\ & \\Leftrightarrow {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}}=12 \\\\ & \\Leftrightarrow {{m}^{2}}-2\\left( m-2 \\right)=12 \\\\ & \\Leftrightarrow {{m}^{2}}-2m-8=0 \\\\ \\end{align}$<br\/>$\\Delta {{'}_{m}}=9\\Rightarrow {{m}_{1}}=1+3=4;{{m}_{2}}=1-3=-2$ <br\/>C\u1ea3 hai gi\u00e1 tr\u1ecb $m_1$ v\u00e0 $m_2$ \u0111\u1ec1u th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n (*)<br\/>V\u1eady $m=4$ ho\u1eb7c $m=-2$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B v\u00e0 D.<\/span><\/span>"}]}],"id_ques":1046},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $m{{x}^{2}}-2\\left( m-1 \\right)x+m=0$ c\u00f3 hai nghi\u1ec7m $x_1;x_2$ th\u1ecfa m\u00e3n $\\dfrac{{{x}_{1}}}{{{x}_{2}}}+\\dfrac{{{x}_{2}}}{{{x}_{1}}}=4$ ","select":["A. $m=\\dfrac{-2\\pm \\sqrt{6}}{2}$ ","B. $m=\\dfrac{2\\pm \\sqrt{6}}{2}$","C. $m=-2\\pm \\sqrt{6}$","D. $m=2\\pm \\sqrt{6}$ "],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $\\Delta '={{\\left( m-1 \\right)}^{2}}-{{m}^{2}}={{m}^{2}}-2m+1-{{m}^{2}}=1-2m$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $x_1$ v\u00e0 $x_2$ $\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & \\Delta '\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & 1-2m\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & m\\le \\dfrac{1}{2} \\\\ \\end{aligned} \\right.\\,\\,\\,\\left( * \\right)$<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=\\dfrac{2\\left( m-1 \\right)}{m} \\\\ & {{x}_{1}}.{{x}_{2}}=1 \\\\ \\end{aligned} \\right.$ <br\/>Ta c\u00f3:<br\/> $\\begin{aligned} & \\dfrac{{{x}_{1}}}{{{x}_{2}}}+\\dfrac{{{x}_{2}}}{{{x}_{1}}}=4\\Leftrightarrow \\dfrac{x_{1}^{2}+x_{2}^{2}}{{{x}_{1}}{{x}_{2}}}=4\\\\ & \\Rightarrow x_{1}^{2}+x_{2}^{2}=4{{x}_{1}}{{x}_{2}} \\\\ & \\Leftrightarrow {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}}=4{{x}_{1}}{{x}_{2}}\\Leftrightarrow {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-6{{x}_{1}}{{x}_{2}}=0 \\\\ & \\Leftrightarrow \\dfrac{4{{\\left( m-1 \\right)}^{2}}}{{{m}^{2}}}-6=0 \\\\ & \\Rightarrow 4\\left( {{m}^{2}}-2m+1 \\right)-6{{m}^{2}}=0\\\\ & \\Leftrightarrow -2{{m}^{2}}-8m+4=0 \\\\ & \\Leftrightarrow {{m}^{2}}+4m-2=0 \\\\ \\end{aligned}$<br\/>${{\\Delta }_{m}}'=4+2=6\\Rightarrow m=-2\\pm \\sqrt{6}$ (th\u1ecfa m\u00e3n(*))<br\/>V\u1eady $m=-2\\pm \\sqrt{6}$ th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n b\u00e0i to\u00e1n<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span><\/span>","column":2}]}],"id_ques":1047},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","3"]],"list":[{"point":5,"img":"","ques":"Gi\u1ea3 s\u1eed ${{x}_{1}};{{x}_{2}}$ l\u00e0 nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh: $m{{x}^{2}}-2\\left( m+2 \\right)x+\\left( m-3 \\right)=0$. T\u00ecm m\u1ed9t h\u1ec7 th\u1ee9c gi\u1eefa $x_1$ v\u00e0 $x_2$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o $m$ ","hint":"","column":1,"number_true":2,"select":["A. $\\dfrac{{{x}_{1}}+{{x}_{2}}-2}{4}=\\dfrac{1-{{x}_{1}}{{x}_{2}}}{3}$","B. $\\dfrac{{{x}_{1}}+{{x}_{2}}-2}{3}=\\dfrac{1-{{x}_{1}}{{x}_{2}}}{4}$","C. $3{{x}_{1}}+3{{x}_{2}}+4{{x}_{1}}{{x}_{2}}-10=0$","D. $3{{x}_{1}}+3{{x}_{2}}+4{{x}_{1}}{{x}_{2}}+10=0$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 nghi\u1ec7m: $\\Delta \\ge 0$ <br\/>B\u01b0\u1edbc 2: Vi\u1ebft h\u1ec7 th\u1ee9c Vi\u2013\u00e9t. T\u1eeb h\u1ec7 th\u1ee9c th\u1ee9 nh\u1ea5t, ta bi\u1ec3u di\u1ec5n $m=f(x_1;x_2)$ v\u00e0 t\u1eeb h\u1ec7 th\u1ee9c th\u1ee9c hai ta bi\u1ec3u di\u1ec5n $m=h(x_1;x_2)$<br\/>B\u01b0\u1edbc 3: T\u1eeb \u0111\u00f3 suy ra $f(x_1;x_2)=h(x_1;x_2).$ \u0110\u00f3 ch\u00ednh l\u00e0 h\u1ec7 th\u1ee9c gi\u1eefa ${{x}_{1}};{{x}_{2}}$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o $m.$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: $\\Delta '={{\\left( m+2 \\right)}^{2}}-m\\left( m-3 \\right)={{m}^{2}}+4m+4-{{m}^{2}}+3m=7m+4$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m $\\Leftrightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & \\Delta '\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & 7m+4\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & m\\ge -\\dfrac{4}{7} \\\\ \\end{aligned} \\right.$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3:<br\/> $\\begin{aligned} & \\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=\\dfrac{2\\left( m+2 \\right)}{m}\\,\\,\\, \\\\ & {{x}_{1}}{{x}_{2}}=\\dfrac{m-3}{m}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=2+\\dfrac{4}{m} \\\\ & {{x}_{1}}{{x}_{2}}=1-\\dfrac{3}{m} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & \\dfrac{4}{m}={{x}_{1}}+{{x}_{2}}-2 \\\\ & \\dfrac{3}{m}=1-{{x}_{1}}{{x}_{2}} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & \\dfrac{1}{m}=\\dfrac{{{x}_{1}}+{{x}_{2}}-2}{4} \\\\ & \\dfrac{1}{m}=\\dfrac{1-{{x}_{1}}{{x}_{2}}}{3} \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow \\dfrac{{{x}_{1}}+{{x}_{2}}-2}{4}=\\dfrac{1-{{x}_{1}}{{x}_{2}}}{3} \\\\ \\end{aligned}$ <br\/>C\u00f3 th\u1ec3 \u0111\u01a1n gi\u1ea3n h\u1ec7 th\u1ee9c thu \u0111\u01b0\u1ee3c nh\u01b0 sau:<br\/> $\\dfrac{{{x}_{1}}+{{x}_{2}}-2}{4}=\\dfrac{1-{{x}_{1}}{{x}_{2}}}{3}\\Leftrightarrow 3\\left( {{x}_{1}}+{{x}_{2}}-2 \\right)=4\\left( 1-{{x}_{1}}{{x}_{2}} \\right)\\Leftrightarrow 3{{x}_{1}}+3{{x}_{2}}+4{{x}_{1}}{{x}_{2}}-10=0$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A v\u00e0 C.<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> C\u00e1c em c\u00f3 th\u1ec3 thay tr\u1ef1c ti\u1ebfp h\u1ec7 th\u1ee9c Vi-\u00e9t v\u00e0o c\u00e1c \u0111\u00e1p \u00e1n tr\u00ean xem h\u1ec7 th\u1ee9c n\u00e0o th\u1ecfa m\u00e3n.<\/span>"}]}],"id_ques":1048},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho parabol (P): $y={{x}^{2}}$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=mx+1$ . T\u00ecm $m$ \u0111\u1ec3 (d) c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t c\u00f3 ho\u00e0nh \u0111\u1ed9 ${{x}_{1}};{{x}_{2}}$ th\u1ecfa m\u00e3n: $x_{1}^{2}{{x}_{2}}+x_{2}^{2}{{x}_{1}}-{{x}_{1}}{{x}_{2}}=3$ <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m=$_input_<\/span>","hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) l\u00e0:<br\/> $\\begin{aligned} & {{x}^{2}}=mx+1 \\\\ & \\Leftrightarrow {{x}^{2}}-mx-1=0\\,\\,\\,\\,\\left( 1 \\right) \\\\ \\end{aligned}$ <br\/>Ta c\u00f3: $\\Delta ={{m}^{2}}+4>0$ v\u1edbi m\u1ecdi $m$<br\/> Suy ra ph\u01b0\u01a1ng tr\u00ecnh (1) lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t.<br\/>Suy ra (d) lu\u00f4n c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=m \\\\ & {{x}_{1}}{{x}_{2}}=-1 \\\\ \\end{aligned} \\right.$<br\/> Theo gi\u1ea3 thi\u1ebft, ta c\u00f3:<br\/> $x_{1}^{2}{{x}_{2}}+x_{2}^{2}{{x}_{1}}-{{x}_{1}}{{x}_{2}}=3$<br\/>$\\begin{aligned} & \\Leftrightarrow {{x}_{1}}{{x}_{2}}\\left( {{x}_{1}}+{{x}_{2}} \\right)-{{x}_{1}}{{x}_{2}}=3 \\\\ & \\Leftrightarrow (-1).m+1=3 \\\\ & \\Leftrightarrow m=-2 \\\\ \\end{aligned}$<br\/> Ta c\u00f3 $m=-2$ th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n b\u00e0i to\u00e1n<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2.$ <\/span><\/span>"}]}],"id_ques":1049},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-1"],["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\left( {{x}^{2}}-x+1 \\right)\\left( {{x}^{2}}-x+2 \\right)-12=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{$_input_;_input_$\\}$<\/span>","hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $t={{x}^{2}}-x+1$ ","explain":"<span class='basic_left'>\u0110\u1eb7t $t={{x}^{2}}-x+1,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,t\\left( t+1 \\right)-12=0 \\\\ & \\Leftrightarrow {{t}^{2}}+t-12=0 \\\\ & \\Delta =49\\Rightarrow \\sqrt{\\Delta }=7 \\\\ & \\Rightarrow {{t}_{1}}=\\dfrac{-1+7}{2}=3;{{t}_{2}}=\\dfrac{-1-7}{2}=-4 \\\\ \\end{align}$ <br\/>+ V\u1edbi $t={{t}_{1}}=3\\Rightarrow {{x}^{2}}-x+1=3\\Leftrightarrow {{x}^{2}}-x-2=0$ <br\/>V\u00ec $a-b+c=1+1-2=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $x=-1;x=2$ <br\/>+ V\u1edbi $t={{t}_{2}}=-4$ , ta c\u00f3 ${{x}^{2}}-x+1=-4\\Leftrightarrow {{x}^{2}}-x+5=0$ <br\/>$\\Delta =-19<0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\left\\{ -1;2 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$ v\u00e0 $2.$ <\/span><\/span>"}]}],"id_ques":1050},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\sqrt{2x+3}+\\sqrt{x+1}=1$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{$_input_$\\}$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 2: V\u00ec hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec1u kh\u00f4ng \u00e2m n\u00ean ta b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf. Bi\u1ebfn \u0111\u1ed5i v\u00e0 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng: $\\sqrt{f(x)}=m$. <br\/>B\u01b0\u1edbc 4: Ti\u1ebfp t\u1ee5c b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf v\u1edbi \u0111i\u1ec1u ki\u1ec7n $m \\ge 0 $ \u0111\u1ec3 t\u00ecm nghi\u1ec7m. So s\u00e1nh nghi\u1ec7m v\u1edbi \u0111i\u1ec1u ki\u1ec7n v\u00e0 k\u1ebft lu\u1eadn. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $\\left\\{ \\begin{aligned} & 2x+3\\ge 0 \\\\ & x+1\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge -\\dfrac{3}{2} \\\\ & x\\ge -1 \\\\ \\end{aligned} \\right.\\Leftrightarrow x\\ge -1$ (*)<br\/>B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh, ta c\u00f3:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,2x+3+2.\\sqrt{2x+3}.\\sqrt{x+1}+x+1=1 \\\\ & \\Leftrightarrow 3x+3+2.\\sqrt{2x+3}.\\sqrt{x+1}=0 \\\\ & \\Leftrightarrow 2.\\sqrt{2x+3}.\\sqrt{x+1}=-3x-3 \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & -3x-3\\ge 0 \\\\ & 4\\left( 2x+3 \\right)\\left( x+1 \\right)={{\\left( -3x-3 \\right)}^{2}} \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x\\le -1 \\\\ & 4\\left( 2{{x}^{2}}+5x+3 \\right)=9{{x}^{2}}+18x+9 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\le -1 \\\\ & {{x}^{2}}-2x-3=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\le -1 \\\\ & \\left[ \\begin{aligned} & x=-1 \\\\ & x=3 \\\\ \\end{aligned} \\right. (\\text{do}\\,a-b+c=0) \\\\ \\end{aligned} \\right.\\Leftrightarrow x=-1$ (th\u1ecfa m\u00e3n (*))<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m l\u00e0 $x=-1$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$<\/span><br\/><b>Ch\u00fa \u00fd:<\/b> Ch\u1ec9 b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf khi c\u1ea3 hai v\u1ebf \u0111\u1ec1u kh\u00f4ng \u00e2m. N\u1ebfu ta kh\u00f4ng \u0111\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 hai v\u1ebf kh\u00f4ng \u00e2m th\u00ec sau khi t\u00ecm \u0111\u01b0\u1ee3c nghi\u1ec7m, ta ph\u1ea3i thay nghi\u1ec7m v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u \u0111\u1ec3 th\u1eed l\u1ea1i.<\/span>"}]}],"id_ques":1051},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $5x-7{{x}^{2}}+8\\sqrt{7{{x}^{2}}-5x+1}=8$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0: ","select":["A. $S=\\left\\{ 0;-\\dfrac{5}{7};3;-\\dfrac{16}{7} \\right\\}$","B. $S=\\left\\{ 0;\\dfrac{5}{7};3;\\dfrac{16}{7} \\right\\}$","C. $S=\\left\\{ 0;\\dfrac{32}{7};6;-\\dfrac{-32}{7} \\right\\}$","D. $S=\\left\\{ 0;\\dfrac{5}{7};3;-\\dfrac{16}{7} \\right\\}$"],"hint":"\u0110\u1eb7t $t=\\sqrt{7{{x}^{2}}-5x+1},t\\ge 0$ ","explain":"<span class='basic_left'><br\/>\u0110i\u1ec1u ki\u1ec7n: $7{{x}^{2}}-5x+1 \\ge 0$ <br\/>\u0110\u1eb7t $t=\\sqrt{7{{x}^{2}}-5x+1},\\,t\\ge 0$ <br\/>Ta c\u00f3 $5x-7{{x}^{2}}=-\\left( 7{{x}^{2}}-5x+1 \\right)+1=-{{t}^{2}}+1$ <br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,-{{t}^{2}}+1+8t=8 \\\\ & \\Leftrightarrow {{t}^{2}}-8t+7=0 \\\\ \\end{aligned}$ <br\/>V\u00ec $a+b+c=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $t=1$ v\u00e0 $t=7$<br\/>C\u1ea3 hai gi\u00e1 tr\u1ecb \u0111\u1ec1u th\u1ecfa m\u00e3n $t\\ge 0$ <br\/>- V\u1edbi $t=1$, ta c\u00f3:<br\/> $\\sqrt{7{{x}^{2}}-5x+1}=1\\Leftrightarrow 7{{x}^{2}}-5x+1=1\\Leftrightarrow 7{{x}^{2}}-5x=0\\Leftrightarrow x\\left( 7x-5 \\right)=0\\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & 7x-5=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=\\dfrac{5}{7} \\\\ \\end{aligned} \\right. (\\text{th\u1ecfa m\u00e3n})$<br\/>- V\u1edbi $t=7,$ ta c\u00f3:<br\/>$\\begin{aligned} & \\sqrt{7{{x}^{2}}-5x+1}=7\\Leftrightarrow 7{{x}^{2}}-5x+1=49\\Leftrightarrow 7{{x}^{2}}-5x-48=0 \\\\ & \\Delta =25+4.7.48=1369\\Rightarrow \\sqrt{\\Delta }=37 \\\\ & \\Rightarrow {{x}_{1}}=\\dfrac{5+37}{14}=3;{{x}_{2}}=\\dfrac{5-37}{14}=-\\dfrac{16}{7}\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned}$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 0;\\dfrac{5}{7};3;-\\dfrac{16}{7} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D. <\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t<\/span><br\/>\u1ede b\u00e0i n\u00e0y, ta kh\u00f4ng n\u00ean c\u00f4 l\u1eadp c\u0103n th\u1ee9c \u1edf m\u1ed9t v\u1ebf r\u1ed3i b\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf v\u00ec khi \u0111\u00f3, qu\u00e1 tr\u00ecnh bi\u1ebfn \u0111\u1ed5i ph\u1ee9c t\u1ea1p h\u01a1n r\u1ea5t nhi\u1ec1u v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh thu \u0111\u01b0\u1ee3c s\u1ebd c\u00f3 b\u1eadc 4.<\/span>","column":2}]}],"id_ques":1052},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{4}}-{{x}^{2}}+2x-1=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0: ","select":["A. $S=\\left\\{ \\dfrac{1+\\sqrt{5}}{2};\\dfrac{1-\\sqrt{5}}{2} \\right\\}$","B. $S=\\left\\{ \\dfrac{-1+\\sqrt{5}}{2};\\dfrac{-1-\\sqrt{5}}{2} \\right\\}$","C. $S=\\left\\{ 1+\\sqrt{5};1-\\sqrt{5} \\right\\}$","D. $S=\\left\\{ -1+\\sqrt{5};-1-\\sqrt{5} \\right\\}$"],"hint":"Nh\u00f3m k\u1ebft h\u1ee3p v\u1edbi s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u1ec3 ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,{{x}^{4}}-{{x}^{2}}+2x-1=0 \\\\ & \\Leftrightarrow {{x}^{4}}-\\left( {{x}^{2}}-2x+1 \\right)=0 \\\\ & \\Leftrightarrow {{x}^{4}}-{{\\left( x-1 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}-x+1 \\right)\\left( {{x}^{2}}+x-1 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}-x+1=0 \\\\ & {{x}^{2}}+x-1=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-x+1=0$ v\u00f4 nghi\u1ec7m v\u00ec $\\Delta =-3<0$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+x-1=0$ c\u00f3 $\\Delta =5>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 $x=\\dfrac{-1\\pm \\sqrt{5}}{2}$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\left\\{ \\dfrac{-1+\\sqrt{5}}{2};\\dfrac{-1-\\sqrt{5}}{2} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span><\/span>","column":2}]}],"id_ques":1053},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{x}{2x-6}+\\dfrac{x}{2x+2}=\\dfrac{2x+4}{{{x}^{2}}-2x-3}$ (1) <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$_input_$\\}$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 2: Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c hai v\u1ebf r\u1ed3i kh\u1eed m\u1eabu th\u1ee9c<br\/>B\u01b0\u1edbc 3: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1eeba nh\u1eadn \u0111\u01b0\u1ee3c<br\/>B\u01b0\u1edbc 4: Trong c\u00e1c gi\u00e1 tr\u1ecb t\u00ecm \u0111\u01b0\u1ee3c c\u1ee7a \u1ea9n, lo\u1ea1i c\u00e1c gi\u00e1 tr\u1ecb kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh. C\u00e1c gi\u00e1 tr\u1ecb th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\left\\{ -1;3 \\right\\}$ <br\/>$\\left( 1 \\right)\\Leftrightarrow \\dfrac{x}{2\\left( x-3 \\right)}+\\dfrac{x}{2\\left( x+1 \\right)}=\\dfrac{2x+4}{\\left( x+1 \\right)\\left( x-3 \\right)}$<br\/>Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,x\\left( x+1 \\right)+x\\left( x-3 \\right)=2\\left( 2x+4 \\right) \\\\ & \\Leftrightarrow {{x}^{2}}+x+{{x}^{2}}-3x=4x+8 \\\\ & \\Leftrightarrow 2{{x}^{2}}-6x-8=0 \\\\ & \\Leftrightarrow {{x}^{2}}-3x-4=0 \\\\ \\end{aligned}$ <br\/>Ta c\u00f3 $a-b+c=1+3-4=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $x=-1$ v\u00e0 $x=4$<br\/>K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n, ta c\u00f3 $x=4$ th\u1ecfa m\u00e3n<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=4$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $4.$ <\/span><\/span>"}]}],"id_ques":1054},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{{{x}^{2}}+x-5}{x}+\\dfrac{3x}{{{x}^{2}}+x-5}+4=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0: ","select":["A. $S=\\left\\{ 1+\\sqrt{6};1-\\sqrt{6};1;-5 \\right\\}$ ","B. $S=\\left\\{ -1+\\sqrt{6};-1-\\sqrt{6};1;-5 \\right\\}$ ","C. $S=\\left\\{ -1+\\sqrt{6};-1-\\sqrt{6};1;5 \\right\\}$ ","D. $S=\\left\\{ -1+\\sqrt{6};-1-\\sqrt{6};-1;5 \\right\\}$ "],"hint":"\u0110\u1eb7t $t=\\dfrac{{{x}^{2}}+x-5}{x}$","explain":"<span class='basic_left'><span class='basic_green'>Ph\u00e2n t\u00edch:<\/span><br\/>\u0110\u00e2y l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu, tuy nhi\u00ean ta kh\u00f4ng n\u00ean gi\u1ea3i theo ph\u01b0\u01a1ng ph\u00e1p th\u00f4ng th\u01b0\u1eddng. V\u00ec n\u1ebfu ta th\u1ef1c hi\u1ec7n quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu, ta thu \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc ba s\u1ebd kh\u00f3 gi\u1ea3i h\u01a1n r\u1ea5t nhi\u1ec1u.<br\/>\u1ede \u0111\u00e2y ta nh\u1eadn th\u1ea5y \u1edf v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh, c\u00f3 2 hai ph\u00e2n th\u1ee9c $\\dfrac{{{x}^{2}}+x-5}{x}$ v\u00e0 $\\dfrac{x}{{{x}^{2}}+x-5}$ l\u00e0 ngh\u1ecbch \u0111\u1ea3o c\u1ee7a nhau. Do \u0111\u00f3 ta d\u00f9ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t \u1ea9n ph\u1ee5 \u0111\u1ec3 quy v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $x\\ne 0$ v\u00e0 ${{x}^{2}}+x-5\\ne 0$ (*)<br\/>\u0110\u1eb7t $t=\\dfrac{{{x}^{2}}+x-5}{x},t\\ne 0$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,t+\\dfrac{3}{t}+4=0 \\\\ & \\Leftrightarrow {{t}^{2}}+4t+3=0 \\\\ \\end{aligned}$ <br\/>V\u00ec $a-b+c=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $t=-1;t=-3$<br\/> + V\u1edbi $t=-1$, ta c\u00f3 <br\/>$\\,\\,\\,\\dfrac{{{x}^{2}}+x-5}{x}=-1\\Rightarrow {{x}^{2}}+x-5=-x\\Leftrightarrow {{x}^{2}}+2x-5=0$<br\/> $\\Delta '=1+5=6\\Rightarrow x=-1\\pm \\sqrt{6}$(th\u1ecfa m\u00e3n (*))<br\/>+ V\u1edbi $t=-3,$ ta c\u00f3:<br\/> $\\,\\,\\,\\,\\,\\dfrac{{{x}^{2}}+x-5}{x}=-3\\Rightarrow {{x}^{2}}+x-5=-3x\\Leftrightarrow {{x}^{2}}+4x-5=0$<br\/> V\u00ec $a+b+c=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 $x=1$ v\u00e0 $x=-5$ (th\u1ecfa m\u00e3n (*))<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ -1+\\sqrt{6};-1-\\sqrt{6};1;-5 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span><\/span>","column":2}]}],"id_ques":1055},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"],["7"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Hai v\u00f2i n\u01b0\u1edbc c\u00f9ng ch\u1ea3y v\u00e0o m\u1ed9t b\u1ec3 th\u00ec sau $2$ gi\u1edd $55$ ph\u00fat b\u1ec3 \u0111\u1ea7y n\u01b0\u1edbc. N\u1ebfu m\u1edf ri\u00eang t\u1eebng v\u00f2i th\u00ec v\u00f2i th\u1ee9 nh\u1ea5t l\u00e0m \u0111\u1ea7y b\u1ec3 nhanh h\u01a1n v\u00f2i th\u1ee9 hai l\u00e0 $2$ gi\u1edd. N\u1ebfu m\u1edf ri\u00eang t\u1eebng v\u00f2i th\u00ec m\u1ed7i v\u00f2i ch\u1ea3y bao l\u00e2u \u0111\u1ea7y b\u1ec3?<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> V\u00f2i I: _input_(gi\u1edd); v\u00f2i II: _input_(gi\u1edd)<\/span>","hint":"","explain":"<span class='basic_left'>\u0110\u1ed5i $2$ gi\u1edd $55$ ph\u00fat$ =2\\dfrac{11}{12}$ (gi\u1edd)$ =\\dfrac{35}{12}$ (gi\u1edd)<br\/>G\u1ecdi $x$ (gi\u1edd) l\u00e0 th\u1eddi gian v\u00f2i I ch\u1ea3y m\u1ed9t m\u00ecnh \u0111\u1ea7y b\u1ec3, $x>0. $<br\/>N\u1ebfu m\u1edf ri\u00eang t\u1eebng v\u00f2i, v\u00f2i th\u1ee9 nh\u1ea5t l\u00e0m \u0111\u1ea7y b\u1ec3 nhanh h\u01a1n v\u00f2i th\u1ee9 hai l\u00e0 $2$ gi\u1edd n\u00ean th\u1eddi gian \u0111\u1ec3 v\u00f2i hai ch\u1ea3y m\u1ed9t m\u00ecnh l\u00e0 $x+2$ (gi\u1edd). <br\/>M\u1ed9t gi\u1edd, v\u00f2i I ch\u1ea3y \u0111\u01b0\u1ee3c $\\dfrac{1}{x}$ (b\u1ec3); v\u00f2i II ch\u1ea3y \u0111\u01b0\u1ee3c $\\dfrac{1}{x+2}$ (b\u1ec3)<br\/>Do hai v\u00f2i n\u01b0\u1edbc c\u00f9ng ch\u1ea3y v\u00e0o b\u1ec3 th\u00ec sau $\\dfrac{35}{12}$ gi\u1edd \u0111\u1ea7y b\u1ec3 n\u00ean $1$ gi\u1edd, hai v\u00f2i ch\u1ea3y \u0111\u01b0\u1ee3c $\\dfrac{12}{35}$ b\u1ec3.<br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\dfrac{1}{x}+\\dfrac{1}{x+2}=\\dfrac{12}{35} \\\\ & \\Rightarrow 35\\left( x+2+x \\right)=12x\\left( x+2 \\right) \\\\ & \\Leftrightarrow 70x+70=12{{x}^{2}}+24x \\\\ & \\Leftrightarrow 12{{x}^{2}}-46x-70=0 \\\\ & \\Leftrightarrow 6{{x}^{2}}-23x-35=0 \\\\ & \\Delta =1369\\Rightarrow \\sqrt{\\Delta }=37 \\\\ & \\Rightarrow {{x}_{1}}=\\dfrac{23+37}{12}=5;{{x}_{2}}=\\dfrac{23-37}{12}=-\\dfrac{7}{6} \\\\ \\end{aligned}$ <br\/>Do $x>0$ n\u00ean $x=x_1=5$<br\/>V\u1eady v\u00f2i I ch\u1ea3y m\u1ed9t m\u00ecnh trong $5$ gi\u1edd th\u00ec \u0111\u1ea7y b\u1ec3; v\u00f2i II ch\u1ea3y m\u1ed9t m\u00ecnh trong $7$ gi\u1edd th\u00ec \u0111\u1ea7y b\u1ec3<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $5;$ $7.$<\/span><\/span>"}]}],"id_ques":1056},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["24"],["30"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"M\u1ed9t m\u1ea3nh v\u01b0\u1eddn h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 di\u1ec7n t\u00edch $720$ $m^2.$ N\u1ebfu t\u0103ng chi\u1ec1u d\u00e0i th\u00eam $10$ $m$ v\u00e0 gi\u1ea3m chi\u1ec1u r\u1ed9ng $6$ $m$ th\u00ec di\u1ec7n t\u00edch m\u1ea3nh v\u01b0\u1eddn kh\u00f4ng \u0111\u1ed5i. T\u00ednh chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng m\u1ea3nh v\u01b0\u1eddn.<br\/><b> \u0110\u00e1p s\u1ed1:<\/b> Chi\u1ec1u r\u1ed9ng: _input_ $(m);$ chi\u1ec1u d\u00e0i: _input_ $(m)$","hint":"","explain":"<span class='basic_left'>G\u1ecdi chi\u1ec1u d\u00e0i c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u l\u00e0 $x$ $(m),$ $x>0$<br\/>V\u00ec di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 $720$ $m^2$ n\u00ean chi\u1ec1u r\u1ed9ng c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 $\\dfrac{720}{x}$ $(m)$<br\/>N\u1ebfu t\u0103ng chi\u1ec1u d\u00e0i th\u00eam $10$ $m$ v\u00e0 gi\u1ea3m chi\u1ec1u r\u1ed9ng \u0111i $6$ $m$ th\u00ec chi\u1ec1u d\u00e0i l\u00e0 $x+10$ $(m), $ chi\u1ec1u r\u1ed9ng l\u00e0 $\\dfrac{720}{x}-6$ $(m).$<br\/> Khi \u0111\u00f3 di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi l\u00e0 $\\left( x+10 \\right)\\left( \\dfrac{720}{x}-6 \\right)$ $(m^2)$<br\/>V\u00ec di\u1ec7n t\u00edch m\u1ea3nh v\u01b0\u1eddn kh\u00f4ng \u0111\u1ed5i n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned}& \\left( x+10 \\right)\\left( \\dfrac{720}{x}-6 \\right)=720 \\\\ & \\Leftrightarrow 720+\\dfrac{7200}{x}-6x-60=720 \\\\ & \\Leftrightarrow \\dfrac{7200}{x}-6x-60=0 \\\\ & \\Leftrightarrow 6{{x}^{2}}+60x-7200=0 \\\\ & \\Leftrightarrow {{x}^{2}}+10x-1200=0 \\\\ & \\Delta '=1225\\Rightarrow \\sqrt{\\Delta '}=35 \\\\ & \\Rightarrow {{x}_{1}}=-5+35=30;{{x}_{2}}=-5-35=-40 \\\\ \\end{aligned}$<br\/> V\u00ec $x>0$ n\u00ean $x=x_1=30$<br\/>V\u1eady chi\u1ec1u d\u00e0i c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 $30$ $m,$ chi\u1ec1u r\u1ed9ng c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 $24$ $m$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $24;$ $30.$ <\/span><\/span>"}]}],"id_ques":1057},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["22"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"M\u1ed9t t\u00e0u tu\u1ea7n tra ch\u1ea1y ng\u01b0\u1ee3c d\u00f2ng $60$ $km,$ sau \u0111\u00f3 ch\u1ea1y xu\u00f4i d\u00f2ng $48$ $km$ tr\u00ean c\u00f9ng d\u00f2ng s\u00f4ng c\u00f3 v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 $2$ $km\/h$. T\u00ednh v\u1eadn t\u1ed1c c\u1ee7a t\u00e0u tu\u1ea7n tra khi n\u01b0\u1edbc y\u00ean l\u1eb7ng, bi\u1ebft th\u1eddi gian xu\u00f4i d\u00f2ng \u00edt h\u01a1n th\u1eddi gian ng\u01b0\u1ee3c d\u00f2ng $1$ gi\u1edd.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b>_input_ $(km\/h)$","hint":"V\u1eadn t\u1ed1c xu\u00f4i d\u00f2ng $=$ v\u1eadn t\u1ed1c th\u1ef1c $+$ v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc<br\/>V\u1eadn t\u1ed1c ng\u01b0\u1ee3c d\u00f2ng $=$ v\u1eadn t\u1ed1c th\u1ef1c $\u2013$ v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>G\u1ecdi v\u1eadn t\u1ed1c th\u1ef1c c\u1ee7a t\u00e0u tu\u1ea7n tra l\u00e0 $x$ $(km\/h)$<br\/>B\u1ea3ng ph\u00e2n t\u00edch chuy\u1ec3n \u0111\u1ed9ng: <br\/> <table><tr><th>C\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng<br><\/th><th>V\u1eadn t\u1ed1c $(km\/h)$<br><\/th><th>Qu\u00e3ng \u0111\u01b0\u1eddng $(km)$<br><\/th><th>Th\u1eddi gian (gi\u1edd)<br><\/th><\/tr><tr><th>Xu\u00f4i d\u00f2ng<br><\/th><td>$x+2$<\/td><td>$48$<\/td><td>$\\dfrac{48}{x+2}$<\/td><\/tr><tr><th>Ng\u01b0\u1ee3c d\u00f2ng<br><\/th><td>$x-2$<\/td><td>$60$<\/td><td>$\\dfrac{60}{x-2}$<\/td><\/tr><\/table><br\/>L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh bi\u1ec3u th\u1ecb th\u1eddi gian xu\u00f4i d\u00f2ng v\u00e0 ng\u01b0\u1ee3c d\u00f2ng c\u1ee7a t\u00e0u.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi v\u1eadn t\u1ed1c c\u1ee7a t\u00e0u tu\u1ea7n tra l\u00e0 $x$ $(km\/h),$ $x>2$<br\/>V\u00ec v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 $2$ $km\/h$ n\u00ean v\u1eadn t\u1ed1c t\u00e0u khi xu\u00f4i d\u00f2ng l\u00e0 $x+2$ $(km\/h),$ v\u1eadn t\u1ed1c t\u00e0u khi ng\u01b0\u1ee3c d\u00f2ng l\u00e0 $x-2$ $(km\/h).$<br\/>Th\u1eddi gian t\u00e0u tu\u1ea7n tra xu\u00f4i d\u00f2ng l\u00e0 $\\dfrac{48}{x+2}$ (gi\u1edd) v\u00e0 th\u1eddi gian khi t\u00e0u tu\u1ea7n tra ng\u01b0\u1ee3c d\u00f2ng l\u00e0 $\\dfrac{60}{x-2}$ (gi\u1edd)<br\/>V\u00ec th\u1eddi gian xu\u00f4i d\u00f2ng \u00edt h\u01a1n th\u1eddi gian ng\u01b0\u1ee3c d\u00f2ng $1$ gi\u1edd n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\dfrac{60}{x-2}-\\dfrac{48}{x+2}=1 \\\\ & \\Rightarrow 60\\left( x+2 \\right)-48\\left( x-2 \\right)=\\left( x+2 \\right)\\left( x-2 \\right) \\\\ & \\Leftrightarrow 12x+216={{x}^{2}}-4 \\\\ & \\Leftrightarrow {{x}^{2}}-12x-220=0 \\\\ & \\Delta '=256\\Rightarrow \\sqrt{\\Delta '}=16 \\\\ & \\Rightarrow {{x}_{1}}=6+16=22;{{x}_{2}}=6-16=-10 \\\\ \\end{aligned}$<br\/> V\u00ec $x>2$ n\u00ean $x=x_1=22$<br\/>V\u1eady v\u1eadn t\u1ed1c c\u1ee7a t\u00e0u tu\u1ea7n tra l\u00e0 $22$ $km\/h.$<br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $22.$<\/span><\/span>"}]}],"id_ques":1058},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $m\\left( m-2 \\right){{x}^{2}}-2mx+3=0$ (1) v\u00f4 nghi\u1ec7m. ","select":["A. $0 \\le m <3$ ","B. $m>3$ ho\u1eb7c $m\\le 0$ ","C. $m>3$ ","D. $m \\ge 0$ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a tham s\u1ed1 \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $ax^2+bx+c=0$ v\u00f4 nghi\u1ec7m<br\/>+ X\u00e9t $a=0,$ khi \u0111\u00f3 ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $bx+c=0:$ N\u1ebfu $b=0,c\\ne 0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>+ X\u00e9t $a\\ne 0$, khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m n\u1ebfu $\\Delta <0$$\\left( \\Delta '<0 \\right)$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>+ X\u00e9t $m\\left( m-2 \\right)=0\\Leftrightarrow \\left[ \\begin{aligned} & m=0 \\\\ & m=2 \\\\ \\end{aligned} \\right.$ <br\/>V\u1edbi $m=0,$ ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh $0x+3=0$ (v\u00f4 l\u00ed) n\u00ean ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m. Suy ra $m=0$ th\u1ecfa m\u00e3n. (*)<br\/>V\u1edbi $m=2,$ ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh $-4x+3=0\\Leftrightarrow x=\\dfrac{3}{4}$<br\/> Suy ra ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=\\dfrac{4}{3}$ n\u00ean $m=2$ kh\u00f4ng th\u1ecfa m\u00e3n \u0111\u1ec1 b\u00e0i<br\/>+ X\u00e9t $m\\left( m-2 \\right)\\ne 0\\Leftrightarrow m\\ne \\left\\{ 0;2 \\right\\}.$ Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m khi $\\,\\Delta '<0$<br\/>$\\begin{aligned} & \\Leftrightarrow {{m}^{2}}-3m\\left( m-2 \\right)<0\\Leftrightarrow {{m}^{2}}-3{{m}^{2}}+6m<0 \\\\ & \\Leftrightarrow -2{{m}^{2}}+6m<0\\Leftrightarrow 2m\\left( m-3 \\right)>0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & \\left\\{ \\begin{aligned} & 2m>0 \\\\ & m-3>0 \\\\ \\end{aligned} \\right. \\\\ & \\left\\{ \\begin{aligned} & 2m<0 \\\\ & m-3<0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & \\left\\{ \\begin{aligned} & m>0 \\\\ & m>3 \\\\ \\end{aligned} \\right. \\\\ & \\left\\{ \\begin{aligned} & m<0 \\\\ & m<3 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & m>3 \\\\ & m<0 \\\\ \\end{aligned} \\right.\\,\\,\\,\\,\\left( ** \\right) \\\\ \\end{aligned}$ <br\/>T\u1eeb (*) v\u00e0 (**), ta c\u00f3 $m>3$ ho\u1eb7c $m\\le 0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span><\/span>","column":2}]}],"id_ques":1059},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["15"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$ cho \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=-x+6$ v\u00e0 parabol (P): $y=x^2. $ G\u1ecdi $A$ v\u00e0 $B$ l\u00e0 hai giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (P). T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $OAB.$<br\/><b> \u0110\u00e1p s\u1ed1:<\/b> Di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ l\u00e0 _input_ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: V\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a (d) v\u00e0 (P). <br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m \u0111\u1ec3 t\u00ecm t\u1ecda \u0111\u1ed9 $A$ v\u00e0 $B$.<br\/>B\u01b0\u1edbc 3: H\u1ea1 $AH\\bot \\text{Ox}$ t\u1ea1i $H,$ $BD$$\\bot $ $Ox$ t\u1ea1i $D$ v\u00e0 $C=\\left( d \\right)\\,\\bigcap \\text{Ox}.$<br\/> T\u1eeb \u0111\u00f3 t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ qua di\u1ec7n t\u00edch $\\Delta{CBD}, \\Delta{OBD},\\Delta{OAC}$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai24/lv2/img\/D24.5.png' \/><\/center><br\/>Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m $A$ v\u00e0 $B$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,{{x}^{2}}=-x+6 \\\\ & \\Leftrightarrow {{x}^{2}}+x-6=0 \\\\ & \\Delta =1+24=25\\Rightarrow \\sqrt{\\Delta }=5 \\\\ & \\Rightarrow {{x}_{A}}=\\dfrac{-1+5}{2}=2;{{x}_{B}}=\\dfrac{-1-5}{2}=-3 \\\\ \\end{align}$<br\/> Thay ${{x}_{A}}=2$ v\u00e0o h\u00e0m s\u1ed1 $y=-x+6,$ ta c\u00f3 ${{y}_{A}}=4$. Suy ra $A(2;4)$<br\/>Thay ${{x}_{B}}=-3$ v\u00e0o h\u00e0m s\u1ed1 $y=-x+6,$ ta c\u00f3 ${{y}_{B}}=9$. Suy ra $B(-3;9)$<br\/>H\u1ea1 $AH\\bot \\text{Ox}$ t\u1ea1i $H,$ $BD$$\\bot $ $Ox$ t\u1ea1i $D$ v\u00e0 $C=\\left( d \\right)\\,\\bigcap \\text{Ox}$<br\/>Suy ra $H(2;0), D(-3;0)$ v\u00e0 $C(6;0)$<br\/>Suy ra $AH=4, BD=9; OC=4$ v\u00e0 $CD=9$<br\/>Ta c\u00f3<br\/>$\\begin{align} & {{S}_{\\vartriangle CBD}}={{S}_{\\vartriangle OBD}}+{{S}_{\\vartriangle OAB}}+{{S}_{\\vartriangle OAC}} \\\\ & \\Rightarrow {{S}_{\\vartriangle OAB}}={{S}_{\\vartriangle CBD}}-{{S}_{\\vartriangle OBD}}-{{S}_{\\vartriangle OAC}} \\\\ \\end{align}$ <br\/>+ ${{S}_{\\Delta CBD}}=\\dfrac{1}{2}BD.DC=\\dfrac{1}{2}.9.9=\\dfrac{81}{2}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>+ ${{S}_{\\Delta OBD}}=\\dfrac{1}{2}BD.DO=\\dfrac{1}{2}.9.3=\\dfrac{27}{2}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>+ ${{S}_{\\Delta OAC}}=\\dfrac{1}{2}AH.OC=\\dfrac{1}{2}.4.6=12$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>Suy ra ${{S}_{\\Delta OAB}}=\\dfrac{81}{2}-\\dfrac{27}{2}-12=15$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $15.$<\/span><\/span>"}]}],"id_ques":1060}],"lesson":{"save":0,"level":2}}