{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-2\\left( m-1 \\right)x+m+1=0$ (1). T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 \u0111\u00fang m\u1ed9t nghi\u1ec7m d\u01b0\u01a1ng","select":["A. $m=3$ ho\u1eb7c $m=-1$ ","B. $m=-3$ ho\u1eb7c $m<-1$","C. $m=3$ ho\u1eb7c $m<-1$"],"hint":"X\u00e9t 3 tr\u01b0\u1eddng h\u1ee3p: ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u; ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p d\u01b0\u01a1ng v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m b\u1eb1ng $0$, nghi\u1ec7m c\u00f2n l\u1ea1i d\u01b0\u01a1ng.","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$\\begin{align} & \\Delta '={{\\left( m-1 \\right)}^{2}}-\\left( m-1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,={{m}^{2}}-3m \\\\ & \\,\\,\\,\\,\\,\\,=m\\left( m-3 \\right) \\\\ \\end{align}$ <br\/>$S=2(m-1); P=m+1$<br\/>C\u00f3 c\u00e1c tr\u01b0\u1eddng h\u1ee3p x\u1ea3y ra:<br\/>+ (1) c\u00f3 nghi\u1ec7m k\u00e9p d\u01b0\u01a1ng: $\\left\\{ \\begin{aligned} & \\Delta '=0 \\\\ & -\\dfrac{b'}{a}>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\left( m-3 \\right)=0 \\\\ & m-1>0 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow\\left\\{ \\begin{aligned} & \\left[ \\begin{aligned} & m=0 \\\\ & m=3 \\\\ \\end{aligned} \\right. \\\\ & m>1 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow m=3$ <br\/>+ (1) c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u $\\Leftrightarrow P<0\\Leftrightarrow m+1<0\\Leftrightarrow m<-1$ <br\/>+ (1) c\u00f3 1 nghi\u1ec7m b\u1eb1ng $0$ v\u00e0 nghi\u1ec7m c\u00f2n l\u1ea1i d\u01b0\u01a1ng:<br\/>(1) c\u00f3 m\u1ed9t nghi\u1ec7m $x=0$ n\u00ean thay $x=0$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), ta c\u00f3: $m+1=0$ $\\Leftrightarrow m=-1$<br\/> Khi \u0111\u00f3 $ \\left( 1 \\right) \\Leftrightarrow {{x}^{2}}+4x=0 \\Leftrightarrow x \\left( x+4 \\right)=0 \\Leftrightarrow \\left[ \\begin{align} & x=0 \\\\ & x=-4<0 \\\\ \\end{align} \\right. $ <br\/>Suy ra tr\u01b0\u1eddng h\u1ee3p n\u00e0y kh\u00f4ng x\u1ea3y ra<br\/>V\u1eady $ m = 3 $ ho\u1eb7c $ m < -1 $ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 \u0111\u00fang 1 nghi\u1ec7m d\u01b0\u01a1ng.<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":3}]}],"id_ques":911},{"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":[[["-9"],["-4"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+2\\left( m+1 \\right)x+{{m}^{2}}+4m+3=0$ v\u1edbi $m$ l\u00e0 tham s\u1ed1. Gi\u1ea3 s\u1eed ${{x}_{1}};{{x}_{2}}$ l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/> \u0110\u1eb7t $A={{x}_{1}}{{x}_{2}}-2\\left( {{x}_{1}}+{{x}_{2}} \\right).$ T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A$ v\u00e0 gi\u00e1 tr\u1ecb t\u01b0\u01a1ng \u1ee9ng c\u1ee7a $m.$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $minA=$ _input_ khi $m=$_input_<\/span> ","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 c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m. <br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a $A$ theo tham s\u1ed1 $m$.<br\/>B\u01b0\u1edbc 3: Vi\u1ebft $A$ d\u01b0\u1edbi d\u1ea1ng $[f(m)]^2+b,$ t\u1eeb \u0111\u00f3 \u0111\u00e1nh gi\u00e1 $A$ \u0111\u1ec3 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $\\Delta '={{\\left( m+1 \\right)}^{2}}-\\left( {{m}^{2}}+4m+3 \\right)=-2m-2$<br\/> Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $\\Leftrightarrow \\Delta '\\ge 0\\Leftrightarrow m\\le -1$ (*)<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi\u2013\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align}& {{x}_{1}}+{{x}_{2}}=-2\\left( m+1 \\right) \\\\ & {{x}_{1}}.{{x}_{2}}={{m}^{2}}+4m+3 \\\\ \\end{align} \\right.$ <br\/>Suy ra:<br\/>$\\begin{align} & A={{m}^{2}}+4m+3+4\\left( m+1 \\right) \\\\ & \\,\\,\\,\\,\\,={{m}^{2}}+4m+3+4m+4 \\\\ & \\,\\,\\,\\,\\,={{m}^{2}}+8m+7 \\\\ & \\,\\,\\,\\,\\,={{\\left( m+4 \\right)}^{2}}-9 \\\\ \\end{align}$ <br\/>V\u00ec ${{\\left( m+4 \\right)}^{2}}\\ge 0$ n\u00ean $A={{\\left( m+4 \\right)}^{2}}-9\\ge -9$ <br\/>D\u1ea5u ''$=$'' x\u1ea3y ra khi v\u00e0 ch\u1ec9 khi $m+4=0$$\\Leftrightarrow m=-4$ (th\u1ecfa m\u00e3n (*))<br\/>V\u1eady gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A$ l\u00e0 $-9$ t\u1ea1i $m=-4$<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 $-9$ v\u00e0 $-4$<\/span><\/span>"}]}],"id_ques":912},{"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":[[["0"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+2mx+m=0$ . T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ sao cho $x_{1}^{2}+x_{2}^{2}$ \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t. <br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $m=$_input_<\/span> ","hint":"T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a $x_{1}^{2}+x_{2}^{2}$ theo tham s\u1ed1 $m$ v\u00e0 \u0111\u01b0a v\u1ec1 d\u1ea1ng: $[f(m)]^2+b.$ Sau \u0111\u00f3 \u0111\u00e1nh gi\u00e1 bi\u1ec3u th\u1ee9c t\u00f9y theo \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $m$ (\u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m)","explain":"<span class='basic_left'>Ta c\u00f3 $\\Delta '={{m}^{2}}-m$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $\\Leftrightarrow \\Delta '\\ge 0\\Leftrightarrow {{m}^{2}}-m\\ge 0\\Leftrightarrow m\\left( m-1 \\right)\\ge 0\\Leftrightarrow \\left[ \\begin{aligned} & m\\ge 1 \\\\ & m\\le 0 \\\\ \\end{aligned} \\right.$ (*)<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=-2m \\\\ & {{x}_{1}}{{x}_{2}}=m \\\\ \\end{aligned} \\right.$ <br\/>Ta c\u00f3:<br\/> $\\begin{aligned} & x_{1}^{2}+x_{2}^{2}={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{\\left( 2m \\right)}^{2}}-2m \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{\\left( 2m \\right)}^{2}}-2.2m.\\dfrac{1}{2}+\\dfrac{1}{4}-\\dfrac{1}{4} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{\\left( 2m-\\dfrac{1}{2} \\right)}^{2}}-\\dfrac{1}{4} \\\\ \\end{aligned}$<br\/>+ N\u1ebfu $m\\ge 1$ th\u00ec $2m-\\dfrac{1}{2}\\ge \\dfrac{3}{2}\\Rightarrow {{\\left( 2m-\\dfrac{1}{2} \\right)}^{2}}-\\dfrac{1}{4}\\ge \\dfrac{9}{4}-\\dfrac{1}{4}=2$ <br\/>V\u1eady $x_{1}^{2}+x_{2}^{2}\\ge 2.$ D\u1ea5u ''$=$'' x\u1ea3y ra khi $m=1$(1)<br\/>+ N\u1ebfu $m\\le 0$ th\u00ec $2m-\\dfrac{1}{2}\\le -\\dfrac{1}{2}\\Rightarrow {{\\left( 2m-\\dfrac{1}{2} \\right)}^{2}}\\ge \\dfrac{1}{4}\\Rightarrow {{\\left( 2m-\\dfrac{1}{2} \\right)}^{2}}-\\dfrac{1}{4}\\ge \\dfrac{1}{4}-\\dfrac{1}{4}=0$<br\/>V\u1eady $x_{1}^{2}+x_{2}^{2}\\ge 0.$ D\u1ea5u ''$=$'' x\u1ea3y ra khi $m=0$(2)<br\/>T\u1eeb (1) v\u00e0 (2), suy ra gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $x_{1}^{2}+x_{2}^{2}$ l\u00e0 $0$, \u0111\u1ea1t \u0111\u01b0\u1ee3c khi $m=0$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0.$<\/span><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span><br\/><b>Sai l\u1ea7m th\u01b0\u1eddng g\u1eb7p: <\/b><br\/>$x_{1}^{2}+x_{2}^{2}={{\\left( 2m-\\dfrac{1}{2} \\right)}^{2}}-\\dfrac{1}{4}\\ge -\\dfrac{1}{4}$<br\/>D\u1ea5u ''$=$'' x\u1ea3y ra khi v\u00e0 ch\u1ec9 khi $2m-\\dfrac{1}{2}=0\\Leftrightarrow m=\\dfrac{1}{4}$<br\/>Tuy nhi\u00ean v\u1edbi $m=\\dfrac{1}{4}$ th\u00ec kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n (*).<\/span>"}]}],"id_ques":913},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: $\\left( a{{x}^{2}}+bx+c \\right)\\left( c{{x}^{2}}+bx+a \\right)=0$ trong \u0111\u00f3 $a,$ $b,$ $c$ l\u00e0 nh\u1eefng s\u1ed1 nguy\u00ean $\\left( a,c\\ne 0 \\right)$ , bi\u1ebft r\u1eb1ng $x={{\\left( \\sqrt{2}+1 \\right)}^{2}}$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean c\u00f3 bao nhi\u00eau nghi\u1ec7m?","select":["A. 1","B. 2","C. 3","D. 4"],"hint":"Kh\u00f4ng m\u1ea5t t\u00ednh t\u1ed5ng qu\u00e1t, gi\u1ea3 s\u1eed $x={{\\left( \\sqrt{2}+1 \\right)}^{2}}=3+2\\sqrt{2}$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $ax^2+bx+c=0.$ T\u1eeb \u0111\u00f3 x\u00e1c \u0111\u1ecbnh m\u1ed1i quan h\u1ec7 gi\u1eefa $a,b,c$. ","explain":"<span class='basic_left'>Kh\u00f4ng m\u1ea5t t\u00ednh t\u1ed5ng qu\u00e1t, gi\u1ea3 s\u1eed $x={{\\left( \\sqrt{2}+1 \\right)}^{2}}=3+2\\sqrt{2}$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $ax^2+bx+c=0$<br\/>$\\Leftrightarrow a{{\\left( 3+2\\sqrt{2} \\right)}^{2}}+b\\left( 3+2\\sqrt{2} \\right)+c=0$<br\/>$\\Leftrightarrow a\\left( 17+12\\sqrt{2} \\right)+3b+2\\sqrt{2}b+c=0$ <br\/>$\\Leftrightarrow \\left( 17a+3b+c \\right)+2\\left( 6a+b \\right)\\sqrt{2}=0$ <br\/>N\u1ebfu $6a+b\\ne 0$ th\u00ec $\\sqrt{2}=-\\dfrac{17a+3b+c}{2\\left( 6a+b \\right)}\\in \\mathbb{Q}$ (v\u00f4 l\u00ed)<br\/>Suy ra $\\left\\{ \\begin{aligned} & 17a+3b+c=0 \\\\ & 6a+b=0 \\\\ \\end{aligned} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{aligned} & 17a-18a+c=0 \\\\ & b=-6a \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & a=c \\\\ & b=-6a \\\\ \\end{aligned} \\right.$ <br\/>Thay $c=a; b=-6a$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho, ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\left( a{{x}^{2}}-6ax+a \\right)\\left( a{{x}^{2}}-6ax+a \\right)=0$ hay $\\left( {{x}^{2}}-6x+1 \\right)\\left( {{x}^{2}}-6x+1 \\right)=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y c\u00f3 hai nghi\u1ec7m l\u00e0 $x=3+2\\sqrt{2}$ v\u00e0 $x=3-2\\sqrt{2}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span>","column":4}]}],"id_ques":914},{"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":[[["1"],["-2"],["1"],["-2"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+ax+b=0$ c\u00f3 hai nghi\u1ec7m $c$ v\u00e0 $d.$ Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+cx+d=0$ c\u00f3 hai nghi\u1ec7m $a$ v\u00e0 $b.$ T\u00ecm $a,$ $b,$ $c,$ $d$ bi\u1ebft c\u00e1c s\u1ed1 \u0111\u00f3 \u0111\u1ec1u kh\u00e1c $0.$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $a=$ _input_; $b=$_input_; $c=$_input_; $d=$_input_<\/span> ","hint":"\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t v\u00e0o hai ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho, t\u1eeb \u0111\u00f3 x\u00e1c \u0111\u1ecbnh $a,b,c,d$","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+ax+b=0,$ ta c\u00f3: $\\left\\{ \\begin{align} & c+d=-a\\,\\,\\,\\,\\left( 1 \\right) \\\\ & cd=b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+cx+d=0,$ ta c\u00f3: $\\left\\{ \\begin{align} & a+b=-c\\,\\,\\,\\,\\left( 3 \\right) \\\\ & ab=d\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 4 \\right) \\\\ \\end{align} \\right.$ <br\/>T\u1eeb (1) suy ra: $a+c=-d.$<br\/> T\u1eeb (3) suy ra: $a+c=-b.$<br\/> Do \u0111\u00f3 $b=d$<br\/>T\u1eeb (2), do $b=d\\ne 0$ n\u00ean $c=1$<br\/>T\u1eeb (4), do $b=d\\ne 0$ n\u00ean $a=1$<br\/>Thay $a=c=1$ v\u00e0o (1), ta \u0111\u01b0\u1ee3c: $d=-2.$ Suy ra $b=-2$<br\/>V\u1eady $a=1;b=-2;c=1;d=-2$<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 $1;-2;1$ v\u00e0 $-2.$ <\/span><\/span>"}]}],"id_ques":915},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+mx+\\left( 2m-4 \\right)=0$ (1) c\u00f3 \u00edt nh\u1ea5t m\u1ed9t nghi\u1ec7m kh\u00f4ng \u00e2m. ","select":["A. $ m > 2 $ ","B. $ m \\ge 2 $","C. $ m \\le 2 $","D. $ m < 2 $"],"hint":"T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $m$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u0111\u1ec1u \u00e2m, t\u1eeb \u0111\u00f3 suy ra \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 \u00edt nh\u1ea5t m\u1ed9t nghi\u1ec7m kh\u00f4ng \u00e2m","explain":"<span class='basic_left'>$\\Delta ={{m}^{2}}-4\\left( 2m-4 \\right)={{\\left( m-4 \\right)}^{2}}\\ge 0$ v\u1edbi m\u1ecdi $m$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 nghi\u1ec7m v\u1edbi m\u1ecdi $m$<br\/> Ta c\u00f3: $S=-m$ v\u00e0 $P=2m-4$<br\/>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u0111\u1ec1u \u00e2m l\u00e0 <br\/>$\\left\\{ \\begin{aligned} & P>0 \\\\ & S<0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & 2m-4>0 \\\\ & -m<0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m>2 \\\\ & m>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow m>2$ <br\/>Suy ra \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 \u00edt nh\u1ea5t m\u1ed9t nghi\u1ec7m kh\u00f4ng \u00e2m l\u00e0 $m\\le 2$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/><b>C\u00e1ch 2:<\/b> Ta chia hai tr\u01b0\u1eddng h\u1ee3p: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3<br\/>+ Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u<br\/>+ Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m c\u00f9ng d\u01b0\u01a1ng<br\/><b>C\u00e1ch 3:<\/b> Ta t\u00ecm nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ theo $m:$<br\/>${{x}_{1}}=\\dfrac{-m-\\left( m-4 \\right)}{2}=2-m;{{x}_{2}}=\\dfrac{-m+\\left( m-4 \\right)}{2}=-2$ <br\/>V\u00ec ${{x}_{2}}=-2$ n\u00ean \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 \u00edt nh\u1ea5t m\u1ed9t nghi\u1ec7m kh\u00f4ng \u00e2m th\u00ec ${{x}_{1}}\\ge 0$ <br\/>$\\Leftrightarrow 2-m\\ge 0\\Leftrightarrow m\\le 2$ <\/span>","column":4}]}],"id_ques":916},{"time":24,"part":[{"time":3,"title":"Cho c\u00e1c s\u1ed1 $a,b,c$ kh\u00e1c nhau \u0111\u00f4i m\u1ed9t, $c\\ne 0$. Bi\u1ebft r\u1eb1ng c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh<br\/>${{x}^{2}}+ax+bc=0\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right)$ v\u00e0 ${{x}^{2}}+bx+ca=0\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right)$<br\/> c\u00f3 \u00edt nh\u1ea5t m\u1ed9t nghi\u1ec7m chung. Ch\u1ee9ng minh r\u1eb1ng c\u00e1c nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a hai ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+cx+ab=0.$ ","title_trans":"H\u00e3y s\u1eafp x\u1ebfp c\u00e1c c\u00e2u sau \u0111\u1ec3 \u0111\u01b0\u1ee3c l\u1eddi gi\u1ea3i \u0111\u00fang","temp":"sequence","correct":[[[2],[6],[5],[1],[3],[4]]],"list":[{"point":10,"image":"img\/1.png","left":["Tr\u1eeb t\u1eebng v\u1ebf hai ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a h\u1ec7 cho nhau, ta \u0111\u01b0\u1ee3c $\\left( a-b \\right){{x}_{0}}=c\\left( a-b \\right).$ Do $a\\ne b$ n\u00ean ${{x}_{0}}=c$ ","V\u1eady hai nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a hai ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+cx+ab=0$","Khi \u0111\u00f3 $x_1, x_2$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}+cx+ab=0$ ","G\u1ecdi ${{x}_{0}}$ l\u00e0 nghi\u1ec7m chung c\u1ee7a (1) v\u00e0 (2). Thay $x=x_0$ v\u00e0o hai ph\u01b0\u01a1ng tr\u00ecnh, ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh: $\\left\\{ \\begin{align} & x_{0}^{2}+a{{x}_{0}}+bc=0 \\\\ & x_{0}^{2}+b{{x}_{0}}+ca=0 \\\\ \\end{align} \\right.$","<span class='basic_left'>G\u1ecdi nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a (1) v\u00e0 (2) theo th\u1ee9 t\u1ef1 l\u00e0 ${{x}_{1}};{{x}_{2}}.$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: ${{x}_{0}}{{x}_{1}}=bc;{{x}_{0}}{{x}_{2}}=ca$<br\/>Do ${{x}_{0}}=c\\ne 0$ n\u00ean ${{x}_{1}}=b;{{x}_{2}}=a$ ","Do ${{x}_{0}}=c;{{x}_{1}}=b$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) n\u00ean $c+b=-a$ hay $a+b=-c.$ Suy ra $x_1+x_2=-c;$ $x_1.x_2=ab$"],"top":115,"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm nghi\u1ec7m chung c\u1ee7a (1) v\u00e0 (2).<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, t\u00ecm hai nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a hai ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 3: L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 hai nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a (1) v\u00e0 (2)<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi ${{x}_{0}}$ l\u00e0 nghi\u1ec7m chung c\u1ee7a (1) v\u00e0 (2). Thay $x=x_0$ v\u00e0o hai ph\u01b0\u01a1ng tr\u00ecnh, ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\left\\{ \\begin{align} & x_{0}^{2}+a{{x}_{0}}+bc=0 \\\\ & x_{0}^{2}+b{{x}_{0}}+ca=0 \\\\ \\end{align} \\right.$<br\/>Tr\u1eeb t\u1eebng v\u1ebf hai ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a h\u1ec7 cho nhau, ta \u0111\u01b0\u1ee3c $\\left( a-b \\right){{x}_{0}}=c\\left( a-b \\right)$ <br\/>Do $a\\ne b$ n\u00ean ${{x}_{0}}=c$ <br\/>G\u1ecdi nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a (1) v\u00e0 (2) theo th\u1ee9 t\u1ef1 l\u00e0 ${{x}_{1}};{{x}_{2}}$ . <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: ${{x}_{0}}{{x}_{1}}=bc;{{x}_{0}}{{x}_{2}}=ca$<br\/>Do ${{x}_{0}}=c\\ne 0$ n\u00ean ${{x}_{1}}=b;{{x}_{2}}=a$ <br\/>Do ${{x}_{0}}=c;{{x}_{1}}=b$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) n\u00ean $c+b=-a$ hay $a+b=-c$<br\/>Suy ra $x_1+x_2=-c;$ $x_1.x_2=ab$<br\/>Khi \u0111\u00f3 $x_1, x_2$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}+cx+ab=0$ <br\/>V\u1eady hai nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a hai ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+cx+ab=0$<\/span>"}]}],"id_ques":917},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-4x+2\\left( m-1 \\right)=0$ (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t nh\u1ecf h\u01a1n $2.$","select":["A. $ 1 < m < \\dfrac{5}{3} $ ","B. $-1 < m < \\dfrac{5}{3} $","C. $ m < -1 $","D. $ m > \\dfrac{5}{3} $"],"hint":"\u0110\u1eb7t $x-2=y.$ Khi \u0111\u00f3 ta \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $x$ v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $y.$ <br\/>V\u1edbi $x<2$ th\u00ec ta c\u00f3 $y<0.$ Do \u0111\u00f3 ta t\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (\u1ea9n $y$) c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t.","explain":"<span class='basic_left'><b>C\u00e1ch 1:<\/b> \u0110\u1eb7t $x-2=y.$ Thay $x=y+2$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh m\u1edbi v\u1edbi \u1ea9n $y:$<br\/>$3{{y}^{2}}+8y+\\left( 2m+2 \\right)=0$ (2)<br\/>$\\Delta' =10-6m;P=\\dfrac{2m+2}{3};S=-\\dfrac{8}{3}$ <br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t nh\u1ecf h\u01a1n $2$ $\\Leftrightarrow $ ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 hai nghi\u1ec7m \u00e2m ph\u00e2n bi\u1ec7t<br\/>$ \\Leftrightarrow \\left\\{ \\begin{aligned} & \\Delta' > 0 \\\\ & P >0 \\\\ & S < 0 \\\\ \\end{aligned} \\right. \\Leftrightarrow \\left \\{ \\begin{aligned} & 10-6m > 0 \\\\ & \\dfrac{2m+2}{3} > 0 \\\\ & -\\dfrac{8}{3} < 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow -1 < m < \\dfrac{5}{3} $ <br\/>V\u1eady v\u1edbi $ -1 < m < \\dfrac{5}{3} $ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t nh\u1ecf h\u01a1n $2.$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B <br\/><\/span><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> \u0110\u00e2y l\u00e0 d\u1ea1ng to\u00e1n c\u00f3 nhi\u1ec1u c\u00e1ch gi\u1ea3i. <br\/><b> C\u00e1ch 2: <\/b><br\/>X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-4x+2\\left( m-1 \\right)=0$(1) <br\/>$\\begin{aligned} & \\Delta '=4-3.2\\left( m-1 \\right)=-6m+10 \\\\ & S=\\dfrac{4}{3};P=\\dfrac{2\\left( m-1 \\right)}{3} \\\\ \\end{aligned}$ <br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t nh\u1ecf h\u01a1n $2$ th\u00ec <br\/>$\\left\\{ \\begin{aligned} & \\Delta '>0 \\\\ & {{x}_{1}}<2 \\\\ & {{x}_{2}}<2 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & \\Delta '>0 \\\\ & {{x}_{1}}-2<0 \\\\ & {{x}_{2}}-2<0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & \\Delta '>0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 3 \\right) \\\\ & \\left( {{x}_{1}}-2 \\right)+\\left( {{x}_{2}}-2 \\right)<0\\,\\,\\,\\,\\,\\,\\,\\,\\left( 4 \\right) \\\\ & \\left( {{x}_{1}}-2 \\right)\\left( {{x}_{2}}-2 \\right)>0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 5 \\right) \\\\ \\end{aligned} \\right.$ <br\/>$\\left( 3 \\right)\\Leftrightarrow -6m+10>0\\Leftrightarrow m<\\dfrac{5}{3}$ <br\/>$\\left( 4 \\right)\\Leftrightarrow {{x}_{1}}{{x}_{2}}-2\\left( {{x}_{1}}+{{x}_{2}} \\right)+4>0\\Leftrightarrow \\dfrac{2\\left( m-1 \\right)}{3}-2\\dfrac{4}{3}+4>0\\Leftrightarrow m>-1$ <br\/>$\\left( 5 \\right)\\Leftrightarrow {{x}_{1}}+{{x}_{2}}-4<0\\Leftrightarrow \\dfrac{4}{3}-4<0$ , lu\u00f4n \u0111\u00fang.<br\/>K\u1ebft h\u1ee3p c\u00e1c k\u1ebft qu\u1ea3, ta \u0111\u01b0\u1ee3c: $ -1 < m < \\dfrac{5}{3} $<br\/><b> C\u00e1ch 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh (1)<br\/>V\u1edbi $\\Delta '>0$ t\u1ee9c $m<\\dfrac{5}{3}$ (*) th\u00ec ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t <br\/>${{x}_{1}}=\\dfrac{2-\\sqrt{10-6m}}{3};{{x}_{1}}=\\dfrac{2+\\sqrt{10-6m}}{3}$ <br\/>Do ${{x}_{1}}>{{x}_{2}}$ n\u00ean \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t nh\u1ecf h\u01a1n $2$ l\u00e0 ${{x}_{2}}<2$ <br\/>$\\Leftrightarrow 2+\\sqrt{10-6m}<6\\Leftrightarrow \\sqrt{10-6m} < 4\\Leftrightarrow 10-6m < 16 \\Leftrightarrow m > -1 $ <br\/>K\u1ebft h\u1ee3p v\u1edbi (*), ta \u0111\u01b0\u1ee3c $ -1 < m < \\dfrac{5}{3} $<\/span>","column":2}]}],"id_ques":918},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $\\left( m-2 \\right){{x}^{2}}+\\left( m-4 \\right)x-2=0$ c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ sao cho $\\left| {{x}_{1}}-{{x}_{2}} \\right|=x_{1}^{2}x_{2}^{2}$ ","select":["A. $m=-1\\pm \\sqrt{5}$ ","B. $m=2\\pm \\sqrt{5}$","C. $m=1\\pm \\sqrt{5}$","D. $m=-2\\pm \\sqrt{5}$"],"hint":"\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta bi\u1ebfn \u0111\u1ed5i h\u1ec7 th\u1ee9c \u0111\u00e3 cho theo $m$, sau \u0111\u00f3 d\u00f9ng \u0111\u1ecbnh ngh\u0129a gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i \u0111\u1ec3 t\u00ecm $m.$","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai l\u00e0 $m-2\\ne 0\\Leftrightarrow m\\ne 2$ <br\/>Khi \u0111\u00f3, v\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $a-b+c=m-2-m+4-2=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}}=-1;{{x}_{2}}=\\dfrac{2}{m-2}$ <br\/>Ta c\u00f3 <br\/>$\\begin{align} & \\left| {{x}_{1}}-{{x}_{2}} \\right|=x_{1}^{2}x_{2}^{2} \\\\ & \\Leftrightarrow \\left| -1-\\dfrac{2}{m-2} \\right|=\\dfrac{4}{{{\\left( m-2 \\right)}^{2}}} \\\\ & \\Leftrightarrow \\left| \\dfrac{m}{m-2} \\right|=\\dfrac{4}{{{\\left( m-2 \\right)}^{2}}}\\,\\,\\,\\,\\,\\left( * \\right) \\\\ \\end{align}$ <br\/>N\u1ebfu $\\dfrac{m}{m-2}\\ge 0\\Leftrightarrow \\left[ \\begin{align} & m>2 \\\\ & m\\le 0 \\\\ \\end{align} \\right.$ th\u00ec <br\/>$\\begin{align} & \\left( * \\right)\\Leftrightarrow \\dfrac{m}{m-2}=\\dfrac{4}{{{\\left( m-2 \\right)}^{2}}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow {{m}^{2}}-2m=4 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow {{\\left( m-1 \\right)}^{2}}=5 \\\\ \\end{align}$ <br\/>$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow m=1\\pm \\sqrt{5}$ (th\u1ecfa m\u00e3n)<br\/> N\u1ebfu $ \\dfrac{m}{m-2} < 0 \\Leftrightarrow 0 < m < 2 $ th\u00ec <br\/> $ \\begin{align} & \\left( * \\right)\\Leftrightarrow \\dfrac{-m}{m-2}=\\dfrac{4}{{{\\left( m-2 \\right)}^{2}}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow 2m-{{m}^{2}}=4 \\\\ \\end{align} $ <br\/> $\\,\\,\\,\\,\\,\\,\\,\\,\\,\\Leftrightarrow {{\\left( m-1 \\right)}^{2}}+3=0 $ (v\u00f4 nghi\u1ec7m)<br\/>V\u1eady gi\u00e1 tr\u1ecb $m$ c\u1ea7n t\u00ecm l\u00e0 $m=1\\pm \\sqrt{5}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":919},{"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":[[["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"Cho parabol $y={{x}^{2}}$. G\u1ecdi $A,$ $B$ l\u00e0 c\u00e1c giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $y=mx+2$ v\u1edbi parabol ($m$ l\u00e0 tham s\u1ed1). T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 \u0111o\u1ea1n th\u1eb3ng $AB$ c\u00f3 \u0111\u1ed9 d\u00e0i nh\u1ecf nh\u1ea5t.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m=$_input_ ","hint":"C\u00f4ng th\u1ee9c t\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111i\u1ec3m $\\left( {{x}_{1}};{{y}_{1}} \\right)$ v\u00e0 $\\left( {{x}_{2}};{{y}_{2}} \\right)$ l\u00e0 $\\sqrt{{{\\left( {{x}_{2}}-{{x}_{1}} \\right)}^{2}}+{{\\left( {{y}_{2}}-{{y}_{1}} \\right)}^{2}}}$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: S\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c t\u00ednh kho\u1ea3ng c\u00e1ch gi\u1eefa hai \u0111i\u1ec3m $\\left( {{x}_{1}};{{y}_{1}} \\right)$ v\u00e0 $\\left( {{x}_{2}};{{y}_{2}} \\right)$ \u0111\u1ec3 t\u00ednh $AB.$ <br\/><b>Ch\u00fa \u00fd:<\/b> Bi\u1ec3u di\u1ec5n $y_1;$ $y_2$ qua $x_1;$ $x_2$ v\u00e0 \u0111\u01b0a bi\u1ec3u th\u1ee9c v\u1ec1 d\u1ea1ng ch\u1ec9 c\u00f3 t\u1ed5ng v\u00e0 t\u00edch c\u1ee7a $x_1$ v\u00e0 $x_2$<br\/>B\u01b0\u1edbc 2: X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a parabol v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng, \u00e1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t \u0111\u1ec3 t\u00ednh $AB^2$ theo $m.$<br\/>B\u01b0\u1edbc 3: T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $AB$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai21/lv3/img\/D21.png' \/><\/center><br\/>G\u1ecdi t\u1ecda \u0111\u1ed9 c\u1ee7a $A$ v\u00e0 $B$ theo th\u1ee9 t\u1ef1 l\u00e0 $\\left( {{x}_{1}};{{y}_{1}} \\right)$ v\u00e0 $\\left( {{x}_{2}};{{y}_{2}} \\right)$.<br\/> C\u00e1c \u0111i\u1ec3m $A,$ $B$ thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $y=mx+2$ n\u00ean ${{y}_{1}}=m{{x}_{1}}+2;{{y}_{2}}=m{{x}_{2}}+2$ <br\/>Ta c\u00f3: <br\/>$\\begin{aligned} & A{{B}^{2}}=\\left( {{x}_{2}}-{{x}_{1}}\\right)^{2} +{{\\left( {{y}_{2}}-{{y}_{1}} \\right)}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( {{x}_{2}}-{{x}_{1}} \\right)^2+{{\\left[ \\left( m{{x}_{2}}+2 \\right)-\\left( m{{x}_{1}}+2 \\right) \\right]}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( 1+{{m}^{2}} \\right){{\\left( {{x}_{2}}-{{x}_{1}} \\right)}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( 1+{{m}^{2}} \\right)\\left[ {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-4{{x}_{1}}{{x}_{2}} \\right] \\\\ \\end{aligned}$ <br\/>Do $A$ v\u00e0 $B$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a parabol $y=x^2$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng $y=mx+2$ n\u00ean $x_1$ v\u00e0 $x_2$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:<br\/> ${{x}^{2}}=mx+2$ hay $x^2-mx-2=0$<br\/>$\\Delta ={{m}^{2}}+8>0$ v\u1edbi m\u1ecdi $m$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t. <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=m \\\\ & {{x}_{1}}{{x}_{2}}=-2 \\\\ \\end{aligned} \\right.$ <br\/>Suy ra $A{{B}^{2}}=\\left( 1+{{m}^{2}} \\right)\\left( {{m}^{2}}+8 \\right)={{m}^{4}}+9{{m}^{2}}+8\\ge 8$ v\u1edbi m\u1ecdi $m$<br\/>Suy ra $AB\\ge 2\\sqrt{2}.$ D\u1ea5u ''$=$'' x\u1ea3y ra khi $m=0$<br\/>V\u1eady $\\min AB=2\\sqrt{2}$ khi $m=0$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0.$<\/span>"}]}],"id_ques":920}],"lesson":{"save":0,"level":3}}