{"segment":[{"time":24,"part":[{"title":"X\u00e9t t\u00ednh \u0111\u00fang sai c\u1ee7a c\u00e1c kh\u1eb3ng \u0111\u1ecbnh. ","title_trans":"","temp":"true_false","correct":[["f","t","f","t","f"]],"list":[{"point":5,"image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["\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 ","N\u1ebfu ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ ($a\\ne 0$ ) c\u00f3 $c<0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t","N\u1ebfu ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ ($a\\ne 0$ ) c\u00f3 $ac\\le 0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m","Ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ ($a\\ne 0$ ), $\\Delta >0$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t ${{x}_{1}}=\\dfrac{-b+\\sqrt{\\Delta }}{2a};{{x}_{2}}=\\dfrac{-b-\\sqrt{\\Delta }}{2a}$ th\u00ec ${{x}_{1}}>{{x}_{2}}$ "],"hint":"","explain":["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 $\\Delta >0,$ t\u1ee9c $ \\left\\{ \\begin{align} & a\\ne 0 \\\\ & \\Delta >0 \\\\ \\end{align} \\right.$ ","<br\/><span class='basic_left'>\u0110\u00fang v\u00ec n\u1ebfu $ac<0$ 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<\/span>","<br\/><span class='basic_left'>Sai v\u00ec ph\u01b0\u01a1ng tr\u00ecnh $-{{x}^{2}}+2x-3=0$ c\u00f3 $c=-3<0$ nh\u01b0ng ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y v\u00f4 nghi\u1ec7m<\/span>","<br\/><span class='basic_left'>\u0110\u00fang v\u00ec n\u1ebfu $ac\\le 0$ m\u00e0 $a\\ne 0$, ta c\u0169ng c\u00f3 $\\Delta \\ge 0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ c\u00f3 nghi\u1ec7m.<\/span>","<br\/><span class='basic_left'>Sai v\u00ec v\u1edbi $a<0$ th\u00ec ${{x}_{1}}<{{x}_{2}}$ v\u00e0 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean ch\u1ec9 \u0111\u00fang khi $a>0$ <\/span>"]}]}],"id_ques":841},{"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":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\left( 2x-1 \\right)\\left( x-2 \\right)=5$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 ","select":["A. $S=\\left\\{ 3;-2 \\right\\}$ ","B. $S=\\left\\{ 3; \\dfrac{1}{2} \\right\\}$","C. $S=\\left\\{3; -\\dfrac{1}{2} \\right\\}$","D. $S=\\left\\{ 2;-3\\right\\}$"],"hint":"Bi\u1ebfn \u0111\u1ed5i \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng t\u1ed5ng qu\u00e1t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{aligned} & \\left( 2x-1 \\right)\\left( x-2 \\right)=5 \\\\ & \\Leftrightarrow 2{{x}^{2}}-4x-x+2=5 \\\\ & \\Leftrightarrow 2{{x}^{2}}-5x-3=0 \\\\ \\end{aligned}$<br\/>Ta c\u00f3 $a=2; b=-5;c=-3$<br\/>$\\Delta ={{\\left( -5 \\right)}^{2}}-4.2.\\left( -3 \\right)=49 >0$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t ${{x}_{1}}=\\dfrac{5+7}{4}=3;{{x}_{2}}=\\dfrac{5-7}{4}=-\\dfrac{1}{2}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":842},{"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: ${{x}^{2}}+\\left( \\sqrt{2}+\\sqrt{3} \\right)x+\\sqrt{6}=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 ","select":["A. $S=\\left\\{ -\\sqrt{2};\\sqrt{3} \\right\\}$ ","B. $S=\\left\\{ -\\sqrt{2};-\\sqrt{3} \\right\\}$","C. $S=\\left\\{ \\sqrt{2};\\sqrt{3} \\right\\}$","D. $S=\\left\\{ \\sqrt{2};-\\sqrt{3} \\right\\}$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai b\u1eb1ng c\u00f4ng th\u1ee9c nghi\u1ec7m:<br\/>B\u01b0\u1edbc 1: X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 $a, b, c$ <br\/>B\u01b0\u1edbc 2: T\u00ednh $\\Delta $ <br\/>B\u01b0\u1edbc 3: \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai \u0111\u1ec3 t\u00ecm nghi\u1ec7m.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3 $a=1;b=\\sqrt{2}+\\sqrt{3}$ ; $c=\\sqrt{6}$<br\/> $\\begin{align} & \\Delta ={{\\left( \\sqrt{2}+\\sqrt{3} \\right)}^{2}}-4\\sqrt{6} \\\\ & \\,\\,\\,\\,\\,=5+2\\sqrt{6}-4\\sqrt{6} \\\\ & \\,\\,\\,\\,\\,=5-2\\sqrt{6} \\\\ & \\,\\,\\,\\,={{\\left( \\sqrt{2}-\\sqrt{3} \\right)}^{2}} \\\\ \\end{align}$ <br\/>Suy ra $\\sqrt{\\Delta }=\\sqrt{3}-\\sqrt{2}$ <br\/>V\u00ec $\\Delta >0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 <br\/>${{x}_{1}}=\\dfrac{-\\left( \\sqrt{2}+\\sqrt{3} \\right)+\\left( \\sqrt{3}-\\sqrt{2} \\right)}{2}=-\\sqrt{2}$ ;<br\/>${{x}_{2}}=\\dfrac{-\\left( \\sqrt{2}+\\sqrt{3} \\right)-\\left( \\sqrt{3}-\\sqrt{2} \\right)}{2}=-\\sqrt{3}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ -\\sqrt{2};-\\sqrt{3} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":843},{"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: $\\left( 5-\\sqrt{2} \\right){{x}^{2}}-10x+5+\\sqrt{2}=0$ <br\/>Gi\u00e1 tr\u1ecb n\u00e0o kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:","select":["A. $\\dfrac{27+10\\sqrt{2}}{23}$ ","B. $\\dfrac{25+10\\sqrt{2}}{23}$","C. $1$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 lo\u1ea1i gi\u00e1 tr\u1ecb kh\u00f4ng ph\u1ea3i l\u00e0 nghi\u1ec7m.","explain":"<span class='basic_left'>Ta c\u00f3 $a=5-\\sqrt{2};b'=-5;c=5+\\sqrt{2}$ <br\/>$\\begin{align}<br\/> & \\Delta '={{5}^{2}}-\\left( 5-\\sqrt{2} \\right)\\left( 5+\\sqrt{2} \\right) \\\\ & \\,\\,\\,\\,\\,=25-\\left[ 25-{{\\left( \\sqrt{2} \\right)}^{2}} \\right] \\\\ & \\,\\,\\,\\,\\,=2 \\\\ \\end{align}$ <br\/>V\u00ec $\\Delta '>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0<br\/>$\\begin{align} & {{x}_{1}}=\\dfrac{5+\\sqrt{2}}{5-\\sqrt{2}}=\\dfrac{{{\\left( 5+\\sqrt{2} \\right)}^{2}}}{25-2}=\\dfrac{27+10\\sqrt{2}}{23}; \\\\ & {{x}_{2}}=\\dfrac{5-\\sqrt{2}}{5-\\sqrt{2}}=1 \\\\ \\end{align}$ <br\/>Suy ra $x=\\dfrac{25+10\\sqrt{2}}{23}$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B <\/span><\/span>","column":3}]}],"id_ques":844},{"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":"C\u1eb7p s\u1ed1 ${{x}_{1}}=2$ v\u00e0 ${{x}_{2}}=5$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ","select":["A.${{x}^{2}}-7x+10=0$ ","B. $\\left( x-2 \\right)\\left( x-5 \\right)=0$ ","C. ${{x}^{2}}+7x+10=0$ ","D. $-{{x}^{2}}+7x-10=0$"],"hint":"C\u1eb7p s\u1ed1 $a;b$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $(x-a)(x-b)=0$","explain":"<span class='basic_left'>C\u1eb7p s\u1ed1 ${{x}_{1}}=2$ v\u00e0 ${{x}_{2}}=5$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>$\\begin{align} & \\left( x-2 \\right)\\left( x-5 \\right)=0 \\\\ & \\Leftrightarrow {{x}^{2}}-7x+10=0 \\\\ & \\Leftrightarrow -{{x}^{2}}+7x-10=0 \\\\ \\end{align}$ <br\/>X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+7x+10=0$ <br\/>$\\begin{aligned} & \\Leftrightarrow \\left( x+2 \\right)\\left( x+5 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x+2=0 \\\\ & x+5=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=-2 \\\\ & x=-5 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Suy ra c\u1eb7p s\u1ed1 ${{x}_{1}}=2$ v\u00e0 ${{x}_{2}}=5$ <b>kh\u00f4ng l\u00e0 <\/b> nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+7x+10=0$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":845},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110i\u1ec1n nghi\u1ec7m d\u01b0\u1edbi d\u1ea1ng s\u1ed1 nguy\u00ean ho\u1eb7c s\u1ed1 th\u1eadp ph\u00e2n","temp":"fill_the_blank_random","correct":[[["0,5"],["-3"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\left( x-3 \\right)\\left( x+3 \\right)+x\\left( x+5 \\right)+6=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_;_input_}","hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/> $\\begin{aligned} & \\left( x-3 \\right)\\left( x+3 \\right)+x\\left( x+5 \\right)+6=0 \\\\ & \\Leftrightarrow {{x}^{2}}-9+{{x}^{2}}+5x+6=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}+5x-3=0 \\\\ \\end{aligned}$<br\/>Ta c\u00f3 $a=2; b=5;c=-3$<br\/>$\\begin{aligned} & \\Delta ={{5}^{2}}-4.2.\\left( -3 \\right)=49>0 \\\\ & \\Rightarrow \\sqrt{\\Delta }=7 \\\\ \\end{aligned}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t ${{x}_{1}}=\\dfrac{-5+7}{4}=\\dfrac{1}{2};{{x}_{2}}=\\dfrac{-5-7}{4}=-3$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ \\dfrac{1}{2};-3 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0,5$ v\u00e0 $-3$ <\/span><\/span>"}]}],"id_ques":846},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{1}{2}$","B. $\\dfrac{1}{3}$","C. $\\dfrac{1}{4}$"],"ques":"Ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}-\\left( m+4 \\right)x+m=0$ c\u00f3 m\u1ed9t nghi\u1ec7m b\u1eb1ng $3.$<br\/>Nghi\u1ec7m c\u00f2n l\u1ea1i l\u00e0 ?","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Thay $x=3$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 t\u00ecm $m$. <br\/>B\u01b0\u1edbc 2: V\u1edbi $m$ v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c, ta gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm nghi\u1ec7m c\u00f2n l\u1ea1i.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}-\\left( m+4 \\right)x+m=0$ c\u00f3 m\u1ed9t nghi\u1ec7m b\u1eb1ng $3$ n\u00ean ta c\u00f3:<br\/>$\\begin{align} & {{2.3}^{2}}-\\left( m+4 \\right).3+m=0 \\\\ & \\Leftrightarrow 18-3m-12+m=0 \\\\ & \\Leftrightarrow -2m+6=0 \\\\ & \\Leftrightarrow m=3 \\\\ \\end{align}$<br\/> V\u1edbi $m=3, $ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: $2{{x}^{2}}-7x+3=0$<br\/> $\\Delta ={{\\left( -7 \\right)}^{2}}-4.2.3=25 >0$<br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0<br\/> ${{x}_{1}}=\\dfrac{7+5}{4}=3;{{x}_{2}}=\\dfrac{7-5}{4}=\\dfrac{1}{2}$ <br\/>V\u1eady nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=\\dfrac{1}{2}$ <\/span>"}]}],"id_ques":847},{"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":"Cho ph\u01b0\u01a1ng tr\u00ecnh: ${{\\left( {{x}^{2}}+2x-5 \\right)}^{2}}={{\\left( {{x}^{2}}-x+5 \\right)}^{2}}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 bao nhi\u00eau nghi\u1ec7m?","select":["A. $0$ ","B. $1$","C. $2$","D. $3$"],"hint":"Chuy\u1ec3n v\u1ebf v\u00e0 \u00e1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & {{\\left( {{x}^{2}}+2x-5 \\right)}^{2}}={{\\left( {{x}^{2}}-x+5 \\right)}^{2}} \\\\ & \\Leftrightarrow {{\\left( {{x}^{2}}+2x-5 \\right)}^{2}}-{{\\left( {{x}^{2}}-x+5 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+2x-5+{{x}^{2}}-x+5 \\right)\\left( {{x}^{2}}+2x-5-{{x}^{2}}+x-5 \\right)=0 \\\\ & \\Leftrightarrow \\left( 2{{x}^{2}}+x \\right)\\left( 3x-10 \\right)=0 \\\\ & \\Leftrightarrow x\\left( 2x+1 \\right)\\left( 3x-10 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & 2x+1=0 \\\\ & 3x-10=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=-\\dfrac{1}{2} \\\\ & x=\\dfrac{10}{3} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{0;-\\dfrac{1}{2};\\dfrac{10}{3}\\right\\}$<br\/>Do \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 3 nghi\u1ec7m <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span><\/span>","column":4}]}],"id_ques":848},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (<,>,=) th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[[">"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'> Khi $a>0$ v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ v\u00f4 nghi\u1ec7m th\u00ec $a{{x}^{2}}+bx+c$ _input_ $0$ v\u1edbi m\u1ecdi $x.$ ","hint":"Bi\u1ebfn \u0111\u1ed5i $a{{x}^{2}}+bx+c$ v\u1ec1 d\u1ea1ng $a{{\\left( x+\\dfrac{b}{2a} \\right)}^{2}}-\\dfrac{{{b}^{2}}-4ac}{4a}$","explain":"<span class='basic_left'>Ta c\u00f3 $a{{x}^{2}}+bx+c=a{{\\left( x+\\dfrac{b}{2a} \\right)}^{2}}-\\dfrac{{{b}^{2}}-4ac}{4a}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ v\u00f4 nghi\u1ec7m th\u00ec $\\Delta <0\\Leftrightarrow {{b}^{2}}-4ac<0$ <br\/>M\u00e0 $a>0$ n\u00ean $\\dfrac{{{b}^{2}}-4ac}{4a}<0\\Leftrightarrow -\\dfrac{{{b}^{2}}-4ac}{4a}>0$ <br\/>L\u1ea1i c\u00f3 $a{{\\left( x+\\dfrac{b}{2a} \\right)}^{2}}\\ge 0$ v\u1edbi m\u1ecdi $a>0$<br\/>N\u00ean $a{{x}^{2}}+bx+c=a{{\\left( x+\\dfrac{b}{2a} \\right)}^{2}}-\\dfrac{{{b}^{2}}-4ac}{4a}>0$ v\u1edbi m\u1ecdi $x.$<br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $>.$ <\/span><\/span>"}]}],"id_ques":849},{"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":"\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $5m{{x}^{2}}-4x-3m=0$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t th\u00ec ","select":["A. $m>0$ ","B. $m<0$ ","C. $m \\ne 0$ ","D. $m=0$ "],"hint":"Ph\u01b0\u01a1ng tr\u00ecnh $ax^2+bx+c=0$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & \\Delta >0 \\\\ \\end{aligned} \\right.$ ","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh $5m{{x}^{2}}-4x-3m=0$ c\u00f3 h\u1ec7 s\u1ed1 $a=5m;b\u2019=-2;c=-3m$<br\/>$\\Delta '={{\\left( -2 \\right)}^{2}}-5m.\\left( -3m \\right)=4+15{{m}^{2}}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh $5m{{x}^{2}}-4x-3m=0$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & \\Delta >0 \\\\ \\end{aligned} \\right.$<br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & 5m\\ne 0 \\\\ & 4+15{{m}^{2}}>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & 4+15{{m}^{2}}>0 \\,\\forall m \\\\ \\end{aligned} \\right.\\Leftrightarrow m\\ne 0$<br\/> V\u1eady $m\\ne 0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><br\/><span class='basic_green'>Sai l\u1ea7m th\u01b0\u1eddng m\u1eafc ph\u1ea3i:<\/span> C\u00e1c em th\u01b0\u1eddng qu\u00ean \u0111i\u1ec1u ki\u1ec7n $a \\ne 0$<\/span>","column":4}]}],"id_ques":850},{"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":"Ph\u01b0\u01a1ng tr\u00ecnh ${{m}^{2}}{{x}^{2}}+mx+4=0$ (1) v\u00f4 nghi\u1ec7m v\u1edbi m\u1ecdi $m \\ne 0,$ <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?","select":["A. \u0110\u00fang","B. Sai"],"hint":"X\u00e9t hai tr\u01b0\u1eddng h\u1ee3p $m=0$ v\u00e0 $m\\ne 0$ ","explain":"<span class='basic_left'>Tr\u01b0\u1eddng h\u1ee3p 1: $m=0.$ Khi \u0111\u00f3 $\\left( 1 \\right)\\Leftrightarrow 4=0$ (kh\u00f4ng t\u1ed3n t\u1ea1i). <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m.<br\/>Tr\u01b0\u1eddng h\u1ee3p 2: $m\\ne 0$. Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m $\\Leftrightarrow \\Delta <0\\Leftrightarrow {{\\left( -m \\right)}^{2}}-4.{{m}^{2}}.4<0\\Leftrightarrow -15{{m}^{2}}<0\\Leftrightarrow m\\ne 0$ <br\/>T\u1eeb hai tr\u01b0\u1eddng h\u1ee3p, ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m v\u1edbi m\u1ecdi $m.$<br\/>Suy ra kh\u1eb3ng \u0111\u1ecbnh b\u00e0i to\u00e1n l\u00e0 sai.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":851},{"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":"Cho ph\u01b0\u01a1ng tr\u00ecnh: $\\left( {{m}^{2}}-m-2 \\right){{x}^{2}}+2\\left( m+1 \\right)x+1=0$ v\u1edbi $m$ l\u00e0 tham s\u1ed1.<br\/>T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ch\u1ec9 c\u00f3 m\u1ed9t ph\u1ea7n t\u1eed<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m=$_input_ ","hint":"T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ch\u1ec9 c\u00f3 m\u1ed9t ph\u1ea7n t\u1eed khi v\u00e0 ch\u1ec9 khi ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 1 nghi\u1ec7m ho\u1eb7c c\u00f3 nghi\u1ec7m k\u00e9p.","explain":"<span class='basic_left'>- X\u00e9t $m^2-m-2=0$<br\/>$\\Leftrightarrow \\left( m+1 \\right)\\left( m-2 \\right)=0\\Leftrightarrow \\left[ \\begin{aligned} & m=-1 \\\\ & m=2 \\\\ \\end{aligned} \\right.$ <br\/>+ V\u1edbi $m=-1,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $0x+1=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>+ V\u1edbi $m=2,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $6x+1=0\\Leftrightarrow x=-\\dfrac{1}{6}$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=-\\dfrac{1}{6}$ <br\/>- X\u00e9t ${{m}^{2}}-m-2\\ne 0\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne -1 \\\\ & m\\ne 2 \\\\ \\end{aligned} \\right.$ <br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p n\u1ebfu $\\Delta '=0\\Leftrightarrow {{\\left( m+1 \\right)}^{2}}-\\left( {{m}^{2}}-m-2 \\right)=0$ <br\/> $\\begin{aligned} & \\Leftrightarrow {{m}^{2}}+2m+1-{{m}^{2}}+m+2=0 \\\\ & \\Leftrightarrow 3m+3=0 \\\\ \\end{aligned}$ <br\/>$\\Leftrightarrow m=-1$ (lo\u1ea1i v\u00ec $m\\ne -1$)<br\/>V\u1eady $m=2$ th\u00ec t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ch\u1ec9 c\u00f3 1 ph\u1ea7n t\u1eed.<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2.$<\/span>"}]}],"id_ques":852},{"time":24,"part":[{"time":3,"title":"Cho ph\u01b0\u01a1ng tr\u00ecnh $m{{x}^{2}}+6\\left( m-2 \\right)x+4m-7=0$ ","title_trans":"N\u1ed1i c\u00e2u \u1edf c\u1ed9t b\u00ean tr\u00e1i v\u1edbi c\u00e2u \u1edf c\u1ed9t b\u00ean ph\u1ea3i \u0111\u1ec3 \u0111\u01b0\u1ee3c kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang","audio":"","temp":"matching","correct":[["3","4","2","1"]],"list":[{"point":5,"image":"","left":["Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p khi","Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t khi ","Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m khi ","Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $x=-\\dfrac{7}{12}$ khi "],"right":["a. $m=0$ ","b. $ \\dfrac{9}{5} < m <4 $ ","c. $m=4$ ho\u1eb7c $m=\\dfrac{9}{5}$ ","d. $m>4$ ho\u1eb7c $m<\\dfrac{9}{5}$v\u00e0 $m \\ne 0$"],"top":60,"hint":"","explain":"<span class='basic_left'> X\u00e9t $m{{x}^{2}}+6\\left( m-2 \\right)x+4m-7=0$ (1)<br\/>Tr\u01b0\u1eddng h\u1ee3p 1: $m=0.$<br\/> Khi \u0111\u00f3: $\\left( 1 \\right)\\Leftrightarrow -12x-7=0\\Leftrightarrow x=-\\dfrac{7}{12}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $x=-\\dfrac{7}{12}$ khi $m=0$<br\/>Tr\u01b0\u1eddng h\u1ee3p 2: $m\\ne 0$ <br\/>$a=m\\ne 0;b'=3\\left( m-2 \\right);c=4m-7$ <br\/>$\\begin{align} & \\Delta '=9{{\\left( m-2 \\right)}^{2}}-m\\left( 4m-7 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,=9\\left( {{m}^{2}}-4m+4 \\right)-4{{m}^{2}}+7m \\\\ & \\,\\,\\,\\,\\,\\,\\,=5{{m}^{2}}-29m+36 \\\\ & \\,\\,\\,\\,\\,\\,\\,=\\left( m-4 \\right)\\left( 5m-9 \\right) \\\\ \\end{align}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p $\\Leftrightarrow \\Delta '=0\\Leftrightarrow \\left[ \\begin{align} & m=4 \\\\ & m=\\dfrac{9}{5} \\\\ \\end{align} \\right.$ (th\u1ecfa m\u00e3n $m\\ne 0$)<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow \\Delta '>0\\Leftrightarrow \\left[ \\begin{aligned} & m>4 \\\\ & m<\\dfrac{9}{5} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & m>4 \\\\ & m<\\dfrac{9}{5}\\,\\,\\,\\,\\,\\,\\left( m\\ne 0 \\right) \\\\ \\end{aligned} \\right.$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m $ \\Leftrightarrow \\Delta '<0 \\Leftrightarrow \\dfrac{9}{5} < m < 4 $ <br\/><span class='basic_pink'>V\u1eady 1 n\u1ed1i v\u1edbi c, 2 n\u1ed1i v\u1edbi d, 3 n\u1ed1i b v\u00e0 4 n\u1ed1i v\u1edbi a.<\/span><\/span>"}]}],"id_ques":853},{"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":"T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh sau c\u00f3 nghi\u1ec7m k\u00e9p: <br\/>${{m}^{2}}{{x}^{2}}-mx-2=0$ ","select":["A. $m=0$","B. $m \\ne 0$","C. V\u1edbi m\u1ecdi $m$","D. Kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $m$"],"hint":"Ph\u01b0\u01a1ng tr\u00ecnh $ ax^2+bx+c=0 $ c\u00f3 nghi\u1ec7m k\u00e9p $\\Leftrightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & \\Delta =0 \\\\ \\end{aligned} \\right.$","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{m}^{2}}{{x}^{2}}-mx-2=0$ c\u00f3 nghi\u1ec7m k\u00e9p $\\Leftrightarrow \\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & \\Delta =0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & {{m}^{2}}\\ne 0 \\\\ & {{m}^{2}}-4.{{m}^{2}}.\\left( -2 \\right)=0 \\\\ \\end{aligned} \\right.$<br\/> $\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & 9{{m}^{2}}=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & m=0 \\\\ \\end{aligned} \\right.$ (kh\u00f4ng x\u1ea3y ra)<br\/>V\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m k\u00e9p.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":854},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-\\left( 3{{m}^{2}}-5m+1 \\right)x-\\left( {{m}^{2}}-4m+5 \\right)=0$ lu\u00f4n c\u00f3 nghi\u1ec7m v\u1edbi m\u1ecdi $m,$ <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?","select":["A. \u0110\u00fang","B. Sai"],"hint":"X\u00e9t d\u1ea5u t\u00edch $ac.$","explain":"<span class='basic_left'><span class='basic_green'>Ph\u01b0\u01a1ng ph\u00e1p ch\u1ee9ng minh ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai $ax^2+bx+c=0$ c\u00f3 nghi\u1ec7m<\/span><br\/>C\u00e1ch 1: Ch\u1ee9ng minh $\\Delta \\ge 0$<br\/>C\u00e1ch 2: Ch\u1ee9ng t\u1ecf $ac<0$.<br\/>\u1ede b\u00e0i to\u00e1n n\u00e0y n\u1ebfu l\u00e0m theo c\u00e1ch 1 th\u00ec t\u00ednh $\\Delta$ kh\u00e1 ph\u1ee9c t\u1ea1p n\u00ean ta th\u1eed d\u00f9ng c\u00e1ch 2.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> X\u00e9t t\u00edch $ac=-(m^2-4m+5)=-(m-2)^2-1<0$ v\u1edbi m\u1ecdi $m$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho lu\u00f4n c\u00f3 nghi\u1ec7m v\u1edbi m\u1ecdi $m$<br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0111\u00fang.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 A<\/span><br\/><span class='basic_green'>Ch\u00fa \u00fd:<\/span><br\/>a. Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 $ac<0$ th\u00ec c\u00f3 nghi\u1ec7m nh\u01b0ng \u0111i\u1ec1u ng\u01b0\u1ee3c l\u1ea1i ch\u01b0a ch\u1eafc \u0111\u00fang.<br\/>b. N\u1ebfu $ac\\le 0$ v\u00e0 $a$ kh\u00e1c $0$ th\u00ec ta c\u00f3 $\\Delta \\ge 0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m v\u1edbi m\u1ecdi $m$ <br\/>c. Ch\u1ec9 v\u1edbi \u0111i\u1ec1u ki\u1ec7n $ac\\le 0,$ ch\u01b0a \u0111\u1ea3m b\u1ea3o ph\u01b0\u01a1ng tr\u00ecnh $ax^2+bx+c=0$ c\u00f3 nghi\u1ec7m. Ch\u1eb3ng h\u1ea1n ph\u01b0\u01a1ng tr\u00ecnh $m^2x^2-mx-2=0$ c\u00f3 $ac=-2{{m}^{2}}\\le 0$ nh\u01b0ng v\u1edbi $m=0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh $0x=2,$ v\u00f4 nghi\u1ec7m. <br\/>Nh\u01b0 v\u1eady khi g\u1eb7p tr\u01b0\u1eddng h\u1ee3p $ac\\le 0,$ ph\u1ea3i x\u00e9t hai tr\u01b0\u1eddng h\u1ee3p $a\\ne 0$ v\u00e0 $a=0$<\/span>","column":2}]}],"id_ques":855},{"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":"Ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-2\\left( a+b+c \\right)x+\\left( ab+bc+ca \\right)=0$ v\u00f4 nghi\u1ec7m v\u1edbi m\u1ecdi $a,b,c,$ <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?","select":["A. \u0110\u00fang","B. Sai"],"hint":"X\u00e9t d\u1ea5u $\\Delta '$","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $a=3;b'=\\left( a+b+c \\right);c=ab+ac+bc$ <br\/>$\\begin{align} & \\Delta '={{\\left( a+b+c \\right)}^{2}}-3\\left( ab+bc+ac \\right) \\\\ & \\,\\,\\,\\,\\,\\,={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \\\\ & \\,\\,\\,\\,\\,\\,=\\dfrac{1}{2}\\left[ \\left( {{a}^{2}}-2ab+{{b}^{2}} \\right)+\\left( {{b}^{2}}-2bc+{{c}^{2}} \\right)+\\left( {{c}^{2}}-2ac+{{c}^{2}} \\right) \\right] \\\\ & \\,\\,\\,\\,\\,\\,=\\dfrac{1}{2}\\left[ {{\\left( a-b \\right)}^{2}}+{{\\left( b-c \\right)}^{2}}+{{\\left( c-a \\right)}^{2}} \\right]\\ge 0 \\\\ \\end{align}$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho lu\u00f4n c\u00f3 nghi\u1ec7m v\u1edbi m\u1ecdi $a,$ $b,$ $c. $<br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh b\u00e0i to\u00e1n l\u00e0 sai<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":856},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["8,65"],["1,35"]]],"list":[{"point":5,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Rada c\u1ee7a m\u1ed9t m\u00e1y bay theo d\u00f5i chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a 1 xe t\u1ea3i trong $10$ ph\u00fat v\u00e0 ph\u00e1t hi\u1ec7n v\u1eadn t\u1ed1c c\u1ee7a xe t\u1ea3i thay \u0111\u1ed5i ph\u1ee5 thu\u1ed9c v\u00e0o th\u1eddi gian b\u1edfi c\u00f4ng th\u1ee9c: <br\/>$v=3{{t}^{2}}-30t+135$ <br\/>($t$ t\u00ednh b\u1eb1ng ph\u00fat, $v$ t\u00ednh b\u1eb1ng $km\/h$).<br\/> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a $t$ khi v\u1eadn t\u1ed1c xe t\u1ea3i l\u00e0 $100km\/h$ (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai).<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $t \\approx$ _input_(ph\u00fat) ho\u1eb7c $t \\approx$ _input_(ph\u00fat)<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: Thay $v=100$ v\u00e0o c\u00f4ng th\u1ee9c v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <br\/>B\u01b0\u1edbc 2: Ki\u1ec3m tra nghi\u1ec7m c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3 $v=3{{t}^{2}}-30t+135$<br\/>V\u00ec v\u1eadn t\u1ed1c xe t\u1ea3i l\u00e0 $100km\/h$ n\u00ean ta c\u00f3:<br\/>$100=3{{t}^{2}}-30t+135\\Leftrightarrow 3{{t}^{2}}-30t+35=0$<br\/> $\\Delta '=15^2-3.35=120\\Rightarrow \\sqrt{\\Delta '}=2\\sqrt{30}$<br\/> Khi \u0111\u00f3 ${{t}_{1}}=\\dfrac{15+2\\sqrt{30}}{3}\\approx 8,65$ ; ${{t}_{2}}=\\dfrac{15-2\\sqrt{30}}{3}\\approx 1,35$<br\/> Do rada theo d\u00f5i trong $10$ ph\u00fat n\u00ean $ 0 < t \\le 10 .$ Do \u0111\u00f3 c\u1ea3 hai gi\u00e1 tr\u1ecb c\u1ee7a $t$ \u0111\u1ec1u th\u00edch h\u1ee3p.<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $8,65$ v\u00e0 $1,35.$ <\/span><\/span>"}]}],"id_ques":857},{"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":"V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $x$ th\u00ec gi\u00e1 tr\u1ecb hai bi\u1ec3u th\u1ee9c sau b\u1eb1ng nhau:<br\/>${{x}^{2}}+2+2\\sqrt{2}$ v\u00e0 $2\\left( 1+\\sqrt{2} \\right)x$","select":["A. $x=-\\sqrt{2}$ ho\u1eb7c $x=2+\\sqrt{2}$","B. $x=\\sqrt{2}$ ho\u1eb7c $x=2-\\sqrt{2}$","C. $x=\\sqrt{2}$ ho\u1eb7c $x=-2-\\sqrt{2}$","D. $x=\\sqrt{2}$ ho\u1eb7c $x=2+\\sqrt{2}$"],"hint":"Cho hai bi\u1ec3u th\u1ee9c b\u1eb1ng nhau sau \u0111\u00f3 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $x$","explain":"<span class='basic_left'>Gi\u00e1 tr\u1ecb hai bi\u1ec3u th\u1ee9c \u0111\u00e3 cho b\u1eb1ng nhau $\\Leftrightarrow {{x}^{2}}+2+2\\sqrt{2}=2\\left( 1+\\sqrt{2} \\right)x$<br\/> $\\Leftrightarrow {{x}^{2}}-2\\left( 1+\\sqrt{2} \\right)x+2+2\\sqrt{2}=0$<br\/> $\\begin{aligned} & \\Delta '={{\\left( 1+\\sqrt{2} \\right)}^{2}}-\\left( 2+2\\sqrt{2} \\right) \\\\ & \\,\\,\\,\\,\\,\\,=3+2\\sqrt{2}-2-2\\sqrt{2} \\\\ & \\,\\,\\,\\,\\,\\,=1 \\\\ \\end{aligned}$<br\/> V\u00ec $\\Delta '>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 <br\/>$\\begin{aligned} & {{x}_{1}}=1+\\sqrt{2}+1=2+\\sqrt{2}; \\\\ & {{x}_{2}}=1+\\sqrt{2}-1=\\sqrt{2} \\\\ \\end{aligned}$<br\/> V\u1eady v\u1edbi $x=\\sqrt{2}$ ho\u1eb7c $x=2+\\sqrt{2}$ th\u00ec gi\u00e1 tr\u1ecb hai bi\u1ec3u th\u1ee9c \u0111\u00e3 cho b\u1eb1ng nhau<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":858},{"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":"<span class='basic_left'>Giao \u0111i\u1ec3m c\u1ee7a hai h\u00e0m s\u1ed1 $y=-\\dfrac{1}{2}{{x}^{2}}$ v\u00e0 $y=x-8$ l\u00e0:<\/span>","select":["<span class='basic_left'>A. $\\left( 1+\\sqrt{17};-8+\\sqrt{17} \\right)$ v\u00e0 $\\left( 1-\\sqrt{17};-8-\\sqrt{17} \\right)$<\/span>","<span class='basic_left'>B. $\\left( -1+\\sqrt{17};-9+\\sqrt{17} \\right)$ v\u00e0 $\\left( -1-\\sqrt{17};-9-\\sqrt{17} \\right)$<\/span>","<span class='basic_left'>C. $\\left( -1+\\sqrt{17};-9-\\sqrt{17} \\right)$ v\u00e0 $\\left( -1-\\sqrt{17};-9+\\sqrt{17} \\right)$<\/span>","<span class='basic_left'>D. $\\left( -1+\\sqrt{17};8+\\sqrt{17} \\right)$ v\u00e0 $\\left( -1-\\sqrt{17};8-\\sqrt{17} \\right)$<\/span>"],"hint":"X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u1ed3 th\u1ecb","explain":"<span class='basic_left'>Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u1ed3 th\u1ecb l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned} & -\\dfrac{1}{2}{{x}^{2}}=x-8 \\\\ & \\Leftrightarrow {{x}^{2}}+2x-16=0 \\\\ \\end{aligned}$<br\/> $\\Delta '=1+16=17$<br\/> Suy ra $x=-1\\pm \\sqrt{17}$<br\/> Thay $x=-1+\\sqrt{17}$ v\u00e0o $y=x-8,$ ta c\u00f3: $y=-9+\\sqrt{17}$<br\/>Thay $x=-1-\\sqrt{17}$ v\u00e0o $y=x-8,$ ta c\u00f3: $y=-9-\\sqrt{17}$<br\/>V\u1eady giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u1ed3 th\u1ecb \u0111\u00e3 cho l\u00e0 $\\left( -1+\\sqrt{17};-9+\\sqrt{17} \\right)$ v\u00e0 $\\left( -1-\\sqrt{17};-9-\\sqrt{17} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><\/span>","column":1}]}],"id_ques":859},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u1eb7p s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"V\u00ed d\u1ee5: B\u1ea1n t\u00ecm \u0111\u01b0\u1ee3c nghi\u1ec7m $(a;b)$ th\u00ec ta \u0111i\u1ec1n $(a;b)$. <br\/>Ch\u00fa \u00fd: \u0110i\u1ec1n c\u1ea3 d\u1ea5u ngo\u1eb7c ''('' ,'')'' v\u00e0 d\u1ea5u ch\u1ea5m ph\u1ea9y '';''","temp":"fill_the_blank_random","correct":[[["(1;3)"],["(5;-5)"]]],"list":[{"point":5,"width":80,"content":"","type_input":"","ques":"Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{align} & 2x+y-5=0 \\\\ & y+{{x}^{2}}=4x \\\\ \\end{align} \\right.$ <br\/>H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $(x;y)$ l\u00e0 _input_ v\u00e0 _input_ ","hint":"D\u00f9ng ph\u01b0\u01a1ng ph\u00e1p th\u1ebf \u0111\u1ec3 gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'>Bi\u1ec3u di\u1ec5n $y$ thep $x$ t\u1eeb ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 nh\u1ea5t, ta \u0111\u01b0\u1ee3c: $y=5-2x$ (1)<br\/>Th\u1ebf (1) v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 hai, ta c\u00f3:<br\/>$\\begin{aligned} & 5-2x+{{x}^{2}}=4x \\\\ & \\Leftrightarrow {{x}^{2}}-6x+5=0 \\\\ & \\Leftrightarrow {{x}^{2}}-x-5x+5=0 \\\\ & \\Leftrightarrow x\\left( x-1 \\right)-5\\left( x-1 \\right)=0 \\\\ & \\Leftrightarrow \\left( x-1 \\right)\\left( x-5 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & x=5 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/> V\u1edbi $x=1$ th\u00ec $\\left( 1 \\right)\\Leftrightarrow y=5-2.1=3$<br\/> V\u1edbi $x=5$ th\u00ec $\\left( 1 \\right)\\Leftrightarrow y=5-2.5=-5$ <br\/>V\u1eady h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m $(x;y)$ l\u00e0 $(1;3)$ v\u00e0 $(5;-5)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $(1;3)$ v\u00e0 $(5;-5)$<\/span><\/span>"}]}],"id_ques":860}],"lesson":{"save":0,"level":2}}