{"segment":[{"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":[[["10"],["0"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Cho parabol (P): $y=x^2,$ \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=4mx+10.$ Gi\u1ea3 s\u1eed (d) lu\u00f4n c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t c\u00f3 ho\u00e0nh \u0111\u1ed9 $x_1$ v\u00e0 $x_2.$ T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $P=x_{1}^{2}+x_{2}^{2}+{{x}_{1}}{{x}_{2}}$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $ min\\,P= $_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: X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (P). T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 (d) lu\u00f4n c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $P$ v\u1ec1 \u1ea9n $m$<br\/>B\u01b0\u1edbc 3: \u0110\u00e1nh gi\u00e1 P \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 ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (P) l\u00e0:<br\/>${{x}^{2}}=4mx+10\\Leftrightarrow {{x}^{2}}-4mx-10=0$ (1)<br\/>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 (d) lu\u00f4n c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t <br\/>$\\Leftrightarrow \\Delta '>0\\Leftrightarrow 4{{m}^{2}}+10>0$ (lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi m)<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=4m \\\\ & {{x}_{1}}{{x}_{2}}=-10 \\\\ \\end{aligned} \\right.$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & P=x_{1}^{2}+x_{2}^{2}+{{x}_{1}}{{x}_{2}} \\\\ & \\,\\,\\,\\,={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-{{x}_{1}}{{x}_{2}} \\\\ & \\,\\,\\,\\,={{\\left( 4m \\right)}^{2}}+10 \\\\ & \\,\\,\\,\\,=16{{m}^{2}}+10 \\\\ \\end{aligned}$<br\/>V\u00ec ${{m}^{2}}\\ge 0\\Rightarrow 16{{m}^{2}}+10\\ge 10$ hay $P\\ge 10$ <br\/>D\u1ea5u \u201c$=$\u201d x\u1ea3y ra khi $m=0$<br\/>V\u1eady gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $P$ l\u00e0 $10$ \u0111\u1ea1t \u0111\u01b0\u1ee3c t\u1ea1i $m=0$<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 $10$ v\u00e0 $0$<\/span><\/span>"}]}],"id_ques":1061},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $x_{1} = -\\dfrac{1}{2}; x_{2}= -2$","B. $x_{1} = 2; x_{2}= \\dfrac{1}{2}$","C. $x_{1} = \\dfrac{1}{2}; x_{2}= -2$"],"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{4}}+3{{x}^{3}}-{{x}^{2}}+3x+2=0$ <br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $ x_1 =$?; $x_2=$?<\/span>","hint":"Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed1i x\u1ee9ng b\u1eadc 4. Ta chia c\u1ea3 hai v\u1ebf cho $x^2$ kh\u00e1c $0,$ r\u1ed3i \u0111\u1eb7t \u1ea9n ph\u1ee5 $t=x+\\dfrac{1}{x}$ \u0111\u1ec3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed1i x\u1ee9ng d\u1ea1ng: $a{{x}^{4}}+b{{x}^{3}}+{{c}^{2}}{{x}^{2}}+ bx+a=0,\\left( a\\ne 0 \\right)$ <br\/><b>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i:<\/b><br\/>V\u00ec $x =0$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00ean chia c\u1ea3 hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh cho $x^2,$ ta \u0111\u01b0\u1ee3c:<br\/>$a\\left( {{x}^{2}}+\\dfrac{1}{{{x}^{2}}} \\right)+b\\left( x+ \\dfrac{1}{x} \\right)+c=0$ <br\/>\u0110\u1eb7t $t=x+ \\dfrac{1}{x}\\Rightarrow {{t}^{2}}={{x}^{2}}+\\dfrac{1}{{{x}^{2}}}+ 2\\Rightarrow {{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{t}^{2}}- 2.$ <br\/>Khi \u0111\u00f3 ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai: $a\\left( {{t}^{2}}-2 \\right)+bt+c=0$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>V\u00ec $x =0$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00ean chia c\u1ea3 hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh cho $x^2,$ ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{align} & 2{{x}^{2}}+3x-1+\\dfrac{3}{x}+\\dfrac{2}{{{x}^{2}}}=0 \\\\ & \\Leftrightarrow 2\\left( {{x}^{2}}+\\dfrac{1}{{{x}^{2}}} \\right)+3\\left( x+\\dfrac{1}{x} \\right)-1=0 \\\\ \\end{align}$<br\/>\u0110\u1eb7t $t=x+\\dfrac{1}{x}\\Rightarrow {{t}^{2}}={{x}^{2}}+\\dfrac{1}{{{x}^{2}}}+2\\Rightarrow {{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{t}^{2}}-2$ <br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai:<br\/>$\\begin{align} & 2\\left( {{t}^{2}}-2 \\right)+3t-1=0 \\\\ & \\Leftrightarrow 2{{t}^{2}}+3t-5=0 \\\\ \\end{align}$ <br\/>Do $a+b+c=2+3-5=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $t=1;t=-\\dfrac{5}{2}$<br\/>+ V\u1edbi $t=1,$ ta c\u00f3 $x+\\dfrac{1}{x}=1\\Leftrightarrow {{x}^{2}}-x+1=0$ (ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m do $\\Delta =-3<0$ )<br\/>+ V\u1edbi $t=-\\dfrac{5}{2},$ ta c\u00f3 $x+\\dfrac{1}{x}=-\\dfrac{5}{2}\\Leftrightarrow 2{{x}^{2}}+5x+2=0$ <br\/>$\\Delta =9>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=\\dfrac{-5+3}{4}=-\\dfrac{1}{2};{{x}_{2}}=\\dfrac{-5-3}{4}=-2$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=\\dfrac{-1}{2};{{x}_{2}}=-2$<\/span>"}]}],"id_ques":1062},{"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":[[["-1"],["3"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{4}}=20x+21$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $S=\\{$_input_;_input_ $\\}$<\/span> ","hint":"\u0110\u01b0a hai v\u1ebf v\u1ec1 l\u0169y th\u1eeba c\u00f9ng b\u1eadc","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\,{{x}^{4}}=20x+21 \\\\ & \\Leftrightarrow {{x}^{4}}+4{{x}^{2}}+4=4{{x}^{2}}+20x+25 \\\\ & \\Leftrightarrow {{\\left( {{x}^{2}}+2 \\right)}^{2}}={{\\left( 2x+5 \\right)}^{2}} \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}+2=2x+5 \\\\ & {{x}^{2}}+2=-2x-5 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}-2x-3=0 \\\\ & {{x}^{2}}+2x+7=0\\,\\,\\,\\left(\\text{ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m v\u00ec}\\,\\Delta'=-6<0 \\right) \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow {{x}^{2}}-2x-3=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-1 \\\\ & x=3 \\\\ \\end{aligned} \\right.\\left(\\text{do}\\, a-b+c=0 \\right) \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ -1;3 \\right\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-1$ v\u00e0 $3.$ <\/span><\/span>"}]}],"id_ques":1063},{"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'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\left( x-1 \\right)\\left( x+1 \\right)\\left( x+3 \\right)\\left( x+5 \\right)=9$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{ 2\\pm \\sqrt{10};-2 \\right\\}$","B. $S=\\left\\{ -2\\pm \\sqrt{10};-2 \\right\\}$","C. $S=\\left\\{ -2\\pm \\sqrt{10};2 \\right\\}$","D. $S=\\left\\{ 2\\pm \\sqrt{10};2 \\right\\}$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc 4 d\u1ea1ng $\\left( x+a \\right)\\left( x+b \\right)\\left( x+c \\right)\\left( x+d \\right)=m$ v\u1edbi $a+b=c+d$<br\/><b>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i:<\/b><br\/>B\u01b0\u1edbc 1: Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0\u1ee3c vi\u1ebft th\u00e0nh: $\\left[ {{x}^{2}}+\\left( a+b \\right)x+ab \\right]\\left[ {{x}^{2}}+\\left( c+d \\right)x+cd \\right]=m.$<br\/> \u0110\u1eb7t $t={{x}^{2}}+\\left( a+b \\right)x,$ ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: $\\left( t+ab \\right)\\left( t+cd \\right)=m$ <br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $t,$ t\u1eeb \u0111\u00f3 t\u00ecm $x$ b\u1eb1ng c\u00e1ch gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+\\left( a+b \\right)x=t$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\left( x-1 \\right)\\left( x+1 \\right)\\left( x+3 \\right)\\left( x+5 \\right)=9 \\\\ & \\Leftrightarrow \\left[ \\left( x-1 \\right)\\left( x+5 \\right) \\right].\\left[ \\left( x+1 \\right)\\left( x+3 \\right) \\right]=9 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+4x-5 \\right)\\left( {{x}^{2}}+4x+3 \\right)=9 \\\\ \\end{aligned}$<br\/>\u0110\u1eb7t $t={{x}^{2}}+4x$, ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh:<br\/> $\\left( t-5 \\right)\\left( t+3 \\right)=9\\Leftrightarrow {{t}^{2}}-2t-24=0$<br\/>$\\Delta '=25>0$ $\\Rightarrow t_1=1+5=6;$ $t_2=1-5=-4$<br\/>+ V\u1edbi $t=6$ $\\Rightarrow {{x}^{2}}+4x-6=0$ <br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ta \u0111\u01b0\u1ee3c $x=-2\\pm \\sqrt{10}$ <br\/>+ V\u1edbi $t=-4$ $\\Rightarrow {{x}^{2}}+4x+4=0\\Leftrightarrow {{\\left( x+2 \\right)}^{2}}=0\\Leftrightarrow x=-2$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ -2\\pm \\sqrt{10};-2 \\right\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":1064},{"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":[[["-3"],["5"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\sqrt{x+4}+\\sqrt{6-x}-\\sqrt{24+2x-{{x}^{2}}}=1$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $S=\\{$_input_;_input_ $\\}$<\/span> ","hint":"Nh\u1eadn th\u1ea5y $24+2x-{x}^{2}=(x+4)(6-x).$ \u0110\u1eb7t $a=\\sqrt{x+4};b=\\sqrt{6-x},$ ta c\u00f3 $a^2+b^2=10$. Ta \u0111i gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi hai \u1ea9n $a$ v\u00e0 $b$","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $\\left\\{ \\begin{aligned} & x+4\\ge 0 \\\\ & 6-x\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow -4\\le x\\le 6$ (*)<br\/>\u0110\u1eb7t $a=\\sqrt{x+4};b=\\sqrt{6-x}$ v\u1edbi $a,b\\ge 0.$ Suy ra $a^2+b^2=10$ (1)<br\/> Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh:<br\/> $\\begin{aligned} & \\,\\,\\,\\,\\,\\,a+b-ab=1\\Leftrightarrow a-1+b\\left( 1-a \\right)=0\\Leftrightarrow \\left( a-1 \\right)\\left( b-1 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & a-1=0 \\\\ & b-1=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & a=1 \\\\ & b=1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>+ Thay $a=1$ v\u00e0o (1), ta c\u00f3: $1+{{b}^{2}}=10\\Leftrightarrow {{b}^{2}}=9\\Leftrightarrow b=3$ (do $b\\ge 0$ )<br\/>Suy ra $\\left\\{ \\begin{aligned} & \\sqrt{x+4}=1 \\\\ & \\sqrt{6-x}=3 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x+4=1 \\\\ & 6-x=9 \\\\ \\end{aligned} \\right.\\Leftrightarrow x=-3$ (th\u1ecfa m\u00e3n (*))<br\/>+ Thay $b=1$ v\u00e0o (1), ta c\u00f3: $1+{{a}^{2}}=10\\Leftrightarrow {{a}^{2}}=9\\Leftrightarrow a=3$ (do $a\\ge 0$ )<br\/>Suy ra $\\left\\{ \\begin{aligned} & \\sqrt{x+4}=3 \\\\ & \\sqrt{6-x}=1 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x+4=9 \\\\ & 6-x=1 \\\\ \\end{aligned} \\right.\\Leftrightarrow x=5$ (th\u1ecfa m\u00e3n (*))<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ -3;5 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-3$ v\u00e0 $5.$ <\/span><\/span>"}]}],"id_ques":1065},{"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'>T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{3}}-m\\left( x+1 \\right)+1=0$ c\u00f3 \u0111\u00fang hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<\/span> ","select":["A. $m<\\dfrac{3}{4}$","B. $m>\\dfrac{3}{4}$","C. $m>\\dfrac{4}{3}$","D. $m<\\dfrac{4}{3}$"],"hint":"Nh\u1eadn th\u1ea5y $a-b+c-d=1-0-m-m-1=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $-1.$ T\u1eeb \u0111\u00f3 ta ph\u00e2n t\u00edch v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i t\u00ecm \u0111i\u1ec1u ki\u1ec7n. ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,\\,{{x}^{3}}-m\\left( x+1 \\right)+1=0\\,\\,\\,\\left( 1 \\right) \\\\ & \\Leftrightarrow {{x}^{3}}+1-m\\left( x+1 \\right)=0 \\\\ & \\Leftrightarrow \\left( x+1 \\right)\\left( {{x}^{2}}-x+1 \\right)-m\\left( x+1 \\right)=0 \\\\ & \\Leftrightarrow \\left( x+1 \\right)\\left( {{x}^{2}}-x+1-m \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x+1=0 \\\\ & {{x}^{2}}-x+1-m=0\\,\\,\\,\\left( * \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 $3$ nghi\u1ec7m ph\u00e2n bi\u1ec7t th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow \\Delta >0$<br\/>$\\Leftrightarrow 1-4\\left( 1-m \\right)>0\\Leftrightarrow 4m-3>0\\Leftrightarrow m>\\dfrac{3}{4}$ <br\/>V\u1eady $m>\\dfrac{3}{4}$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 ba nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B. <\/span><\/span>","column":2}]}],"id_ques":1066},{"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":[[["40"],["50"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>H\u00e0 N\u1ed9i c\u00e1ch Nam \u0110\u1ecbnh $90$ $km.$ Hai \u00f4 t\u00f4 kh\u1edfi h\u00e0nh \u0111\u1ed3ng th\u1eddi, xe th\u1ee9 nh\u1ea5t \u0111i t\u1eeb H\u00e0 N\u1ed9i, xe th\u1ee9 hai \u0111i t\u1eeb Nam \u0110\u1ecbnh v\u00e0 ng\u01b0\u1ee3c chi\u1ec1u nhau. Sau $1$ gi\u1edd ch\u00fang g\u1eb7p nhau. Ti\u1ebfp t\u1ee5c \u0111i, xe th\u1ee9 hai t\u1edbi H\u00e0 N\u1ed9i tr\u01b0\u1edbc khi xe th\u1ee9 nh\u1ea5t t\u1edbi Nam \u0110\u1ecbnh l\u00e0 $27$ ph\u00fat. T\u00ednh v\u1eadn t\u1ed1c m\u1ed7i xe.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b>Xe th\u1ee9 nh\u1ea5t: _input_ $(km\/h)$, xe th\u1ee9 hai: _input_ $(km\/h)$<\/span>","hint":"V\u00ec sau $1$ gi\u1edd, hai xe g\u1eb7p nhau n\u00ean t\u1ed5ng qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c c\u1ee7a hai xe trong $1$ gi\u1edd b\u1eb1ng t\u1ed5ng v\u1eadn t\u1ed1c c\u1ee7a hai xe","explain":"<span class='basic_left'>G\u1ecdi v\u1eadn t\u1ed1c c\u1ee7a xe th\u1ee9 nh\u1ea5t (\u0111i t\u1eeb H\u00e0 N\u1ed9i) l\u00e0 $x$ $(km\/h),$ $ 0 < x < 90.$ <br\/>V\u00ec sau $1$ gi\u1edd, hai xe g\u1eb7p nhau n\u00ean t\u1ed5ng qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c c\u1ee7a hai xe trong $1$ gi\u1edd b\u1eb1ng t\u1ed5ng v\u1eadn t\u1ed1c c\u1ee7a hai xe, t\u1ee9c $90$ $km\/h.$ <br\/>Do \u0111\u00f3 v\u1eadn t\u1ed1c c\u1ee7a xe th\u1ee9 hai l\u00e0 $90-x$ (km\/h)<br\/>Qu\u00e3ng \u0111\u01b0\u1eddng m\u00e0 xe th\u1ee9 nh\u1ea5t \u0111i ti\u1ebfp l\u00e0 $90-x$ $(km)$ n\u00ean th\u1eddi gian \u0111\u1ec3 xe th\u1ee9 nh\u1ea5t \u0111i ti\u1ebfp \u0111\u1ec3 t\u1edbi Nam \u0110\u1ecbnh l\u00e0 $\\dfrac{90-x}{x}$ (gi\u1edd)<br\/>Qu\u00e3ng \u0111\u01b0\u1eddng m\u00e0 xe th\u1ee9 hai \u0111i ti\u1ebfp l\u00e0 $x$ $(km) $ n\u00ean th\u1eddi gian xe th\u1ee9 hai \u0111i ti\u1ebfp \u0111\u1ec3 t\u1edbi H\u00e0 N\u1ed9i l\u00e0 $\\dfrac{x}{90-x}$ (gi\u1edd)<br\/>Sau khi ti\u1ebfp t\u1ee5c \u0111i, xe th\u1ee9 hai t\u1edbi H\u00e0 N\u1ed9i tr\u01b0\u1edbc khi xe th\u1ee9 nh\u1ea5t t\u1edbi Nam \u0110\u1ecbnh l\u00e0 $27$ ph\u00fat$=\\dfrac{9}{20}$ (gi\u1edd) n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\dfrac{90-x}{x}-\\dfrac{x}{90-x}=\\dfrac{9}{20} \\\\ & \\Rightarrow 20{{\\left( 90-x \\right)}^{2}}-20{{x}^{2}}=9x\\left( 90-x \\right) \\\\ & \\Leftrightarrow 20\\left( {{90}^{2}}-2.90x+{{x}^{2}} \\right)-20{{x}^{2}}=810x-9{{x}^{2}} \\\\ & \\Leftrightarrow 9{{x}^{2}}-4410x+162000=0 \\\\ & \\Leftrightarrow {{x}^{2}}-490x+18000=0 \\\\ \\end{align}$ <br\/>$\\Delta '=42025\\Rightarrow \\sqrt{\\Delta '}=205$ n\u00ean ${{x}_{1}}=245+205=450;{{x}_{2}}=245-205=40$ <br\/>V\u00ec $ 0 < x < 90 $ n\u00ean ch\u1ec9 c\u00f3 $x_1=40$ th\u1ecfa m\u00e3n. <br\/>V\u1eady v\u1eadn t\u1ed1c c\u1ee7a xe th\u1ee9 nh\u1ea5t l\u00e0 $40$ $km\/h;$ v\u1eadn t\u1ed1c c\u1ee7a xe th\u1ee9 hai l\u00e0 $50$ $km\/h.$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $40$ v\u00e0 $50.$ <\/span><\/span>"}]}],"id_ques":1067},{"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"],["6"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","ques":"<span class='basic_left'>\u0110\u1ec3 ch\u00e0o m\u1eebng ng\u00e0y Gi\u1ea3i ph\u00f3ng mi\u1ec1n Nam 30 th\u00e1ng 4, hai ph\u00e2n x\u01b0\u1edfng c\u01a1 kh\u00ed thi \u0111ua s\u1ea3n xu\u1ea5t. M\u1ed7i ph\u00e2n x\u01b0\u1edfng ph\u1ea3i l\u00e0m $240$ s\u1ea3n ph\u1ea9m trong m\u1ed9t th\u1eddi gian nh\u1ea5t \u0111\u1ecbnh. M\u1ed7i ng\u00e0y ph\u00e2n x\u01b0\u1edfng I s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n ph\u00e2n x\u01b0\u1edfng II l\u00e0 $8$ s\u1ea3n ph\u1ea9m v\u00e0 \u0111\u00e3 ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c s\u1edbm h\u01a1n th\u1eddi gian quy \u0111\u1ecbnh l\u00e0 $3$ ng\u00e0y v\u00e0 s\u1edbm h\u01a1n ph\u00e2n x\u01b0\u1edfng II l\u00e0 $1$ ng\u00e0y. H\u1ecfi th\u1eddi gian ph\u00e2n x\u01b0\u1edfng I v\u00e0 ph\u00e2n x\u01b0\u1edfng II ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c l\u00e0 bao nhi\u00eau?<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> Ph\u00e2n x\u01b0\u1edfng I: _input_(ng\u00e0y), ph\u00e2n x\u01b0\u1edfng II: _input_ (ng\u00e0y)<\/span> ","hint":"G\u1ecdi \u1ea9n l\u00e0 th\u1eddi gian quy \u0111\u1ecbnh \u0111\u1ec3 ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c","explain":"<span class='basic_left'>G\u1ecdi th\u1eddi gian quy \u0111\u1ecbnh l\u00e0 $x$ (ng\u00e0y), $x>0$<br\/>V\u00ec ph\u00e2n x\u01b0\u1edfng I l\u00e0m xong tr\u01b0\u1edbc th\u1eddi h\u1ea1n $3$ ng\u00e0y n\u00ean th\u1ef1c t\u1ebf h\u1ecd ch\u1ec9 l\u00e0m trong $x-3$ (ng\u00e0y).<br\/>V\u00ec ph\u00e2n x\u01b0\u1edfng II xong sau ph\u00e2n x\u01b0\u1edfng I l\u00e0 $1$ ng\u00e0y n\u00ean th\u1ef1c t\u1ebf h\u1ecd ch\u1ec9 l\u00e0m trong $x-2$ (ng\u00e0y)<br\/>Do \u0111\u00f3 m\u1ed7i ng\u00e0y ph\u00e2n x\u01b0\u1edfng I l\u00e0 \u0111\u01b0\u1ee3c $\\dfrac{240}{x-3}$ (s\u1ea3n ph\u1ea9m), ph\u00e2n x\u01b0\u1edfng II l\u00e0m \u0111\u01b0\u1ee3c $\\dfrac{240}{x-2}$ (s\u1ea3n ph\u1ea9m)<br\/>V\u00ec m\u1ed7i ng\u00e0y ph\u00e2n x\u01b0\u1edfng I l\u00e0m \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n ph\u00e2n x\u01b0\u1edfng II l\u00e0 $8$ s\u1ea3n ph\u1ea9m n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{align}<br\/> & \\,\\,\\,\\,\\dfrac{240}{x-3}-\\dfrac{240}{x-2}=8 \\\\ & \\Rightarrow 240\\left( x-2 \\right)-240\\left( x-3 \\right)=8\\left( x-2 \\right)\\left( x-3 \\right) \\\\ & \\Leftrightarrow 240x-480-240x+720=8\\left( {{x}^{2}}-5x+6 \\right) \\\\ & \\Leftrightarrow 8{{x}^{2}}-40x-192=0 \\\\ & \\Leftrightarrow {{x}^{2}}-5x-24=0 \\\\ \\end{align}$<br\/> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $\\Delta =121$ n\u00ean ${{x}_{1}}=\\dfrac{5+11}{2}=8;{{x}_{2}}=\\dfrac{5-11}{2}=-3$<br\/> V\u00ec $x>0$ n\u00ean $x=x_1=8$<br\/>V\u1eady th\u1eddi gian ph\u00e2n x\u01b0\u1edfng I ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c l\u00e0 $x-3=8-3=5$ (ng\u00e0y), th\u1eddi gian ph\u00e2n x\u01b0\u1edfng II ho\u00e0n th\u00e0nh c\u00f4ng vi\u1ec7c l\u00e0 $x-2=8-2=6$ (ng\u00e0y) <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 $5;$ $6.$<\/span><\/span>"}]}],"id_ques":1068},{"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":"<span class='basic_left'>Trong m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 cho \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=mx+1$ v\u00e0 parabol (P): $y=x^2.$ T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $m$ \u0111\u1ec3 (d) c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t $A,$ $B$ sao cho di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ b\u1eb1ng $2$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","select":["A. $m=\\pm \\sqrt{3}$","B. $m=\\pm 5\\sqrt{3}$","C. $m=\\pm 2\\sqrt{3}$ ","D. $m=\\pm 3\\sqrt{3}$"],"hint":" Ch\u1ec9 ra $OA$ v\u00e0 $OB$ vu\u00f4ng g\u00f3c v\u1edbi nhau b\u1eb1ng c\u00e1ch ch\u1ee9ng minh t\u00edch h\u1ec7 s\u1ed1 g\u00f3c b\u1eb1ng $-1$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (P). Nh\u1eadn x\u00e9t v\u1ec1 d\u1ea5u hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/>B\u01b0\u1edbc 2: Bi\u1ec3u di\u1ec5n t\u1ecda \u0111\u1ed9 hai \u0111i\u1ec3m $A$ v\u00e0 $B$. T\u00ecm h\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $OA$ v\u00e0 $OB$. Ch\u1ee9ng minh $OA$ vu\u00f4ng g\u00f3c v\u1edbi $OB$<br\/>B\u01b0\u1edbc 3: T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ v\u00e0 \u00e1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t \u0111\u01b0a v\u1ec1 bi\u1ec3u th\u1ee9c ch\u1ec9 ch\u1ee9a \u1ea9n $m$<br\/>B\u01b0\u1edbc 4: D\u1ef1a v\u00e0o gi\u1ea3 thi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ b\u1eb1ng $2,$ ta t\u00ecm $m$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai24/lv3/img\/K24.1.png' \/><\/center><br\/>Ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (P) l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:<br\/>${{x}^{2}}=mx+1\\Leftrightarrow {{x}^{2}}-mx-1=0$ (1)<br\/>Do $ ac=-1 < 0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh (1) lu\u00f4n c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u $x_1 < 0 < x_2 $<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=m \\\\ & {{x}_{1}}{{x}_{2}}=-1 \\\\ \\end{align} \\right.$<br\/>Gi\u1ea3 s\u1eed $A(x_1;y_1)$ v\u00e0 $B(x_2;y_2).$<br\/> Do $A$ v\u00e0 $B $ thu\u1ed9c (P) n\u00ean ${{y}_{1}}=x_{1}^{2};{{y}_{2}}=x_{2}^{2}$ <br\/>C\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $OA$ v\u00e0 $OB$ l\u1ea7n l\u01b0\u1ee3t c\u00f3 h\u1ec7 s\u1ed1 g\u00f3c l\u00e0 ${{k}_{1}}=\\dfrac{{{y}_{1}}}{{{x}_{1}}}={{x}_{1}};{{k}_{2}}=\\dfrac{{{y}_{2}}}{{{x}_{2}}}={{x}_{2}}$ <br\/>$\\begin{align} & OA=\\sqrt{x_{1}^{2}+y_{1}^{2}}=\\sqrt{x_{1}^{2}+x_{1}^{4}}=\\sqrt{x_{1}^{2}\\left( 1+x_{1}^{2} \\right)}; \\\\ & OB=\\sqrt{x_{2}^{2}+y_{2}^{2}}=\\sqrt{x_{2}^{2}+x_{2}^{4}}=\\sqrt{x_{2}^{2}\\left( 1+x_{2}^{2} \\right)} \\\\ \\end{align}$ <br\/>V\u00ec $k_1.k_2=x_1.x_2=-1$ n\u00ean $OA$ vu\u00f4ng g\u00f3c v\u1edbi $OB$<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $OAB$ l\u00e0:<br\/> $\\begin{align} & \\dfrac{1}{2}.OA.OB=\\dfrac{1}{2}\\sqrt{x_{1}^{2}\\left( 1+x_{1}^{2} \\right)}.\\sqrt{x_{2}^{2}\\left( 1+x_{2}^{2} \\right)}\\,=\\dfrac{1}{2}\\sqrt{x_{1}^{2}x_{2}^{2}\\left( 1+x_{1}^{2} \\right)\\left( 1+x_{2}^{2} \\right)} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{1}{2}\\sqrt{2+x_{1}^{2}+x_{2}^{2}}=4=\\,\\,\\dfrac{1}{2}\\sqrt{2+{{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}}}=\\dfrac{1}{2}\\sqrt{{{m}^{2}}+4} \\\\ \\end{align}$. <br\/>V\u00ec di\u1ec7n t\u00edch tam gi\u00e1c OAB b\u1eb1ng $2$ n\u00ean ta c\u00f3 $\\dfrac{1}{2}\\sqrt{4+{{m}^{2}}}=2\\Leftrightarrow 4+{{m}^{2}}=16\\Leftrightarrow {{m}^{2}}=12\\Leftrightarrow m=\\pm 2\\sqrt{3}$ <br\/>V\u1eady $m=\\pm 2\\sqrt{3}$ th\u00ec th\u1ecfa m\u00e3n b\u00e0i to\u00e1n.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><br\/><span class='basic_green'>Ghi nh\u1edb:<\/span><br\/> H\u1ec7 s\u1ed1 g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $y=ax+b$ l\u00e0 $a=tg\\alpha$ trong \u0111\u00f3 $\\alpha$ l\u00e0 g\u00f3c t\u1ea1o b\u1edfi \u0111\u01b0\u1eddng th\u1eb3ng $y=ax+b$ v\u00e0 tr\u1ee5c $Ox$<\/span>","column":2}]}],"id_ques":1069},{"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":[[["-10"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $A=3\\left( \\dfrac{{{a}^{2}}}{{{b}^{2}}}+\\dfrac{{{b}^{2}}}{{{a}^{2}}} \\right)-8\\left( \\dfrac{a}{b}+\\dfrac{b}{a} \\right)$ <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $min\\,A=$_input_ ","hint":" \u0110\u1eb7t $m=\\dfrac{a}{b}+\\dfrac{b}{a}.$ Ch\u1ee9ng minh $m^2 \\ge 4$<br\/>\u0110\u01b0a $A$ v\u1ec1 d\u1ea1ng ${{\\left[ f\\left( m \\right) \\right]}^{2}}+b$ r\u1ed3i \u0111\u00e1nh gi\u00e1 $A$ v\u1edbi $m\\ge 2$ v\u00e0 $m\\le -2$ ","explain":"<span class='basic_left'>\u0110\u1eb7t $m=\\dfrac{a}{b}+\\dfrac{b}{a}$ <br\/>Ta c\u00f3: ${{m}^{2}}={{\\left( \\dfrac{a}{b}+\\dfrac{b}{a} \\right)}^{2}}=\\dfrac{{{a}^{2}}}{{{b}^{2}}}+\\dfrac{{{b}^{2}}}{{{a}^{2}}}+2\\ge 2\\dfrac{a}{b}.\\dfrac{b}{a}+2=4$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c C\u00f4-si)<br\/>Suy ra ${{m}^{2}}\\ge 4\\Leftrightarrow \\left| m \\right|\\ge 2\\Leftrightarrow \\left[ \\begin{aligned} & m\\ge 2 \\\\ & m\\le -2 \\\\ \\end{aligned} \\right.$<br\/> $A=3\\left[ {{\\left( \\dfrac{a}{b}+\\dfrac{b}{a} \\right)}^{2}}-2 \\right]-8\\left( \\dfrac{a}{b}+\\dfrac{b}{a} \\right)=3\\left( {{m}^{2}}-2 \\right)-8m=3{{m}^{2}}-8m-6$<br\/>Ta c\u00f3: $A=3{{\\left( m-\\dfrac{4}{3} \\right)}^{2}}-\\dfrac{34}{3}$ <br\/>V\u1edbi $m\\ge 2$ ta c\u00f3: <br\/>$\\begin{aligned} & m-\\dfrac{4}{3}\\ge 2-\\dfrac{4}{3}=\\dfrac{2}{3}\\Rightarrow {{\\left( m-\\dfrac{4}{3} \\right)}^{2}}\\ge \\dfrac{4}{9} \\\\ & \\Rightarrow A=3{{\\left( m-\\dfrac{4}{3} \\right)}^{2}}-\\dfrac{34}{3}=3.\\dfrac{4}{9}-\\dfrac{34}{3}=-10\\,\\,\\,\\left( 1 \\right) \\\\ \\end{aligned}$<br\/>D\u1ea5u ''$=$'' x\u1ea3y ra khi $m=2$<br\/>V\u1edbi $m\\le -2$ ta c\u00f3: <br\/>$\\begin{aligned} & m-\\dfrac{4}{3}\\le -2-\\dfrac{4}{3}=-\\dfrac{10}{3}\\Rightarrow {{\\left( m-\\dfrac{4}{3} \\right)}^{2}}\\ge \\dfrac{100}{9} \\\\ & \\Rightarrow A=3{{\\left( m-\\dfrac{4}{3} \\right)}^{2}}-\\dfrac{34}{3}=3.\\dfrac{100}{9}-\\dfrac{34}{3}=22\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned}$<br\/>D\u1ea5u ''$=$'' x\u1ea3y ra khi $m=-2$<br\/>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 $\\min A=-10$ v\u1edbi $m=2,$ khi \u0111\u00f3 $a=b$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $-10$<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>+ Sai l\u1ea7m c\u1ea7n tr\u00e1nh: Ta \u0111\u00e1nh gi\u00e1 $A$ lu\u00f4n nh\u01b0 sau m\u00e0 kh\u00f4ng x\u00e9t c\u00e1c tr\u01b0\u1eddng h\u1ee3p c\u1ee7a $m$:<br\/>V\u00ec ${{\\left( m-\\dfrac{4}{3} \\right)}^{2}}\\ge 0$ n\u00ean $A=3{{\\left( m-\\dfrac{4}{3} \\right)}^{2}}-\\dfrac{34}{3}\\ge -\\dfrac{34}{3}$<br\/>D\u1ea5u ''$=$'' x\u1ea3y ra $\\Leftrightarrow m=\\dfrac{4}{3}$ (kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n (*))<br\/>+ N\u1ebfu ta kh\u00f4ng mu\u1ed1n x\u00e9t t\u1eebng tr\u01b0\u1eddng h\u1ee3p c\u1ee7a $m$ th\u00ec ta ph\u1ea3i bi\u1ebfn \u0111\u1ed5i $A$ sao cho khi $A$ \u0111\u1ea1t gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t th\u00ec $m$ lu\u00f4n t\u1ed3n t\u1ea1i.<br\/> V\u00ed d\u1ee5 nh\u01b0 c\u00e1ch gi\u1ea3i sau:<br\/>$\\begin{aligned}& A=3{{m}^{2}}-8m-6=\\left( {{m}^{2}}-4 \\right)+\\left( 2{{m}^{2}}-8m+8 \\right)-10 \\\\ & \\,\\,\\,\\,\\,=\\left( {{m}^{2}}-4 \\right)+2{{\\left( m-2 \\right)}^{2}}-10 \\\\ \\end{aligned}$<br\/> Do $ {{m}^{2}} \\ge 4,$ ${{\\left( m-2 \\right)}^{2}} \\ge 0 $ n\u00ean $ A \\ge -10 $ <br\/>D\u1ea5u $=$ x\u1ea3y ra $\\Leftrightarrow \\left\\{ \\begin{align} & {{m}^{2}}=4 \\\\ & m-2=0 \\\\ \\end{align} \\right.\\Leftrightarrow m=2.$ Khi \u0111\u00f3 $a=b$<br\/>V\u1eady $\\min A=-10$ khi $a=b$<\/span>"}]}],"id_ques":1070}],"lesson":{"save":0,"level":3}}