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{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $-\\sqrt{2016}$","B.$-2\\sqrt{2016}$","C. $2\\sqrt{2016}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/5.jpg' \/><\/center>Nghi\u1ec7m nh\u1ecf nh\u1ea5t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $(x^2-2014)(x^2-2015)(x^2-2016)=0$ l\u00e0: <br\/>$x=?$<br\/><span class='basic_left'><\/span>","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00edch v\u00e0 so s\u00e1nh nghi\u1ec7m.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & ({{x}^{2}}-2014)({{x}^{2}}-2015)({{x}^{2}}-2016)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}-2014=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(1) \\\\ & {{x}^{2}}-2015=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(2) \\\\ & {{x}^{2}}-2016=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(3) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Gi\u1ea3i (1)<br\/>$\\begin{aligned}& {{x}^{2}}-2014=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\sqrt{2014} \\\\ & x=-\\sqrt{2014} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Gi\u1ea3i (2)<br\/>$\\begin{aligned}& {{x}^{2}}-2015=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\sqrt{2015} \\\\ & x=-\\sqrt{2015} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Gi\u1ea3i (3):<br\/>$\\begin{aligned}& {{x}^{2}}-2016=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\sqrt{2016} \\\\ & x=-\\sqrt{2016} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady nghi\u1ec7m nh\u1ecf nh\u1ea5t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=-\\sqrt {2016}$<\/span>"}]}],"id_ques":881},{"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"]]],"list":[{"point":10,"width":50,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/8.jpg' \/><\/center>Nghi\u1ec7m nh\u1ecf nh\u1ea5t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $(x^2+2017)(x^3-x)=0$ l\u00e0: $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & ({{x}^{2}}+2017)({{x}^{3}}-x)=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+2017 \\right)\\left[ x\\left( {{x}^{2}}-1 \\right) \\right]=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+2017 \\right)\\left[ x\\left( x-1 \\right)\\left( x+1 \\right) \\right]=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}+2017=0 \\\\ & x=0 \\\\ & x-1=0 \\\\ & x+1=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}=-2017\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m v\u00ec } x\\ge 0 \\forall x) \\\\ & x=0 \\\\ & x=1 \\\\ & x=-1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady nghi\u1ec7m nh\u1ecf nh\u1ea5t c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=-1$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$<\/span><\/span>"}]}],"id_ques":882},{"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":[[["-4"],["-1","2"],["2","-1"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/9.jpg' \/><\/center><span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh $x^3+x^2+mx-4=0$. (1)<br\/>a. T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=-2$. <br\/>b. V\u1edbi gi\u00e1 tr\u1ecb $m$ t\u00ecm \u0111\u01b0\u1ee3c \u1edf c\u00e2u a, t\u00ecm c\u00e1c nghi\u1ec7m c\u00f2n l\u1ea1i.<br\/><b>\u0110\u00e1p s\u1ed1: <\/b>$a. m=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\\\ b. x\\in\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span>","hint":"V\u00ec $x=-2$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00ean ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 m\u1ed9t nh\u00e2n t\u1eed $x+2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Thay gi\u00e1 tr\u1ecb c\u1ee7a $x$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1) \u0111\u1ec3 t\u00ecm $m$<br\/><b>B\u01b0\u1edbc 2:<\/b> Thay gi\u00e1 tr\u1ecb $m$ t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 t\u00ecm $x$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><b>a.<\/b> Thay $x=-2$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), ta c\u00f3:<br\/>$(-2)^3+(-2)^2+m(-2)-4=0\\Leftrightarrow -2m-8 \\Leftrightarrow m=-4$<br\/>V\u1eady v\u1edbi $m=-4$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=-2$<br\/><b>b. <\/b>V\u1edbi $m=-4$, ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh: $x^3+x^2-4x-4=0$ (2)<br\/>V\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=-2$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nh\u00e2n t\u1eed $x+2$<br\/>Khi \u0111\u00f3, ta c\u00f3:<br\/>$x^3+x^2-4x-4=0\\\\ \\Leftrightarrow x^3+2x^2-x^2-4x-4=0\\\\ \\Leftrightarrow x^2(x+2)-(x+2)^2=0\\\\ \\Leftrightarrow (x+2)[x^2-(x+2)]=0\\\\ \\Leftrightarrow (x+2)(x^2-x-2)=0\\\\ \\Leftrightarrow \\left[\\begin {aligned}&x+2=0\\\\&x^2-x-2=0 \\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin {aligned}&x+2=0\\\\&x^2+x-2x-2=0 \\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin {aligned}&x+2=0\\\\&x(x+1)-2(x+1)=0 \\\\\\end{aligned}\\right. \\\\ \\Leftrightarrow \\left[\\begin {aligned}&x=-2\\\\&(x+1)(x-2)=0 \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin {aligned}&x=-2\\\\&x=-1\\\\&x=2\\\\ \\end{aligned}\\right.$<br\/>V\u1eady c\u00e1c nghi\u1ec7m c\u00f2n l\u1ea1i c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x=2$ v\u00e0 $x=-1$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-4$; $2;-1$<\/span><br\/><i><span class='basic_green'>Ghi nh\u1edb:<\/span><\/i> Ph\u01b0\u01a1ng tr\u00ecnh $f(x)=0$ (trong \u0111\u00f3, $f(x)$ l\u00e0 m\u1ed9t \u0111a th\u1ee9c) c\u00f3 m\u1ed9t nghi\u1ec7m $x=a$ t\u1ee9c l\u00e0 $f(a)=0$ th\u00ec \u0111a th\u1ee9c $f(x)$ c\u00f3 m\u1ed9t nh\u00e2n t\u1eed l\u00e0 $x-a$. <br\/>Hay<br\/>Ta c\u00f3: $f(x)=0 \\Leftrightarrow (x-a)g(x)=0$. (trong \u0111\u00f3 $g(x)$ l\u00e0 \u0111a th\u1ee9c c\u00f3 b\u1eadc nh\u1ecf h\u01a1n $f(x)$)<br\/>Ta c\u00f3 th\u1ec3 x\u00e1c \u0111\u1ecbnh \u0111a th\u1ee9c $g(x)$ b\u1eb1ng c\u00e1ch th\u1ef1c hi\u1ec7n ph\u00e9p chia \u0111a th\u1ee9c $f(x)$ cho $x-a$.<\/span>"}]}],"id_ques":883},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/3.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x^3-5x^2+8x-4=0$.<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{1;2\\}$","B. $S=\\{-1;-2\\}$","C. $S=\\{-1\\}$","D. \u0110\u00e1p \u00e1n kh\u00e1c"],"hint":"V\u00ec $1+(-5)+8+(-4)=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $1$ nghi\u1ec7m l\u00e0 $x = 1$.","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b>B\u01b0\u1edbc 1:<\/b> S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch $-5x^2=-x^2-4x^2$<br\/><b>B\u01b0\u1edbc 2:<\/b> S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p nh\u00f3m<br\/><b>B\u01b0\u1edbc 3:<\/b> Bi\u1ebfn \u0111\u1ed5i v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed trung<br\/><b>B\u01b0\u1edbc 4:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00edch.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Nh\u1eadn th\u1ea5y: $x=1$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00ean ta c\u00f3 c\u00e1ch gi\u1ea3i sau:<br\/>Ta c\u00f3:<br\/>$x^3-5x^2+8x-4=0\\\\ \\Leftrightarrow x^3-x^2-4x^2+8x-4=0\\\\ \\Leftrightarrow (x^3-x^2)-(4x^2-8x+4)=0\\\\ \\Leftrightarrow x^2(x-1)-4(x^2+2x+1)=0\\\\ \\Leftrightarrow x^2(x-1)-4(x-1)^2=0\\\\ \\Leftrightarrow (x-1)[x^2-4(x-1)]=0\\\\ \\Leftrightarrow (x-1) (x^2-4x+4)=0\\\\ \\Leftrightarrow \\left[ \\begin{aligned}&x-1=0 \\\\ & x^2-4x+4=0\\\\ \\end{aligned}\\right. \\\\ \\Leftrightarrow \\left[ \\begin{aligned}&x=1 \\\\ & (x-2)^2=0\\\\ \\end{aligned}\\right. \\\\ \\Leftrightarrow \\left[ \\begin{aligned}&x=1 \\\\ & x=2\\\\ \\end{aligned}\\right.$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{1;2\\}$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><br\/><b> Nh\u1eadn x\u00e9t:<\/b> Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 d\u1ea1ng $ax^3 + bx^2 + cx + d = 0$ v\u1edbi $a + b + c + d = 0$ th\u00ec c\u00f3 $1$ nghi\u1ec7m l\u00e0 $x = 1$<br\/>T\u1ed5ng qu\u00e1t: Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1ed5ng h\u1ec7 s\u1ed1 b\u1eb1ng $0$ th\u00ec c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=1$. <\/span>","column":2}]}],"id_ques":884},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/11.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x^3-x^2-x-2=0$.<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{-1;2\\}$","B. $S=\\{2\\}$","C. $S=\\{-1\\}$","D. \u0110\u00e1p \u00e1n kh\u00e1c"],"hint":"Nh\u1ea9m t\u00ecm nghi\u1ec7m nguy\u00ean c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. T\u1eeb \u0111\u00f3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch r\u1ed3i gi\u1ea3i.","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Nh\u1eadn th\u1ea5y $x=2$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/>Ta \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u v\u1ec1 d\u1ea1ng: $(x-2).g(x)=0$, trong \u0111\u00f3: $(x-2).g(x)=x^3-x^2-x-2$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/>$x^3-x^2-x-2=0\\\\ \\Leftrightarrow x^3-2x^2+x^2-2x+x-2=0\\\\ \\Leftrightarrow x^2(x-2)+x(x-2)+(x-2)=0 \\\\ \\Leftrightarrow (x-2)(x^2+x+1)=0 \\\\ \\Leftrightarrow \\left[ \\begin{aligned} & x-2=0\\\\ &x^2+x+1=0 \\\\ \\end{aligned}\\right. \\\\ \\Leftrightarrow \\left[ \\begin{aligned} & x=2\\\\ &\\left(x+\\dfrac {1}{2}\\right)^2+\\dfrac {3}{4}=0\\,\\,\\,\\text{(v\u00f4 nghi\u1ec7m v\u00ec }\\left(x+\\dfrac {1}{2}\\right)^2\\ge 0 \\,\\forall x\\,\\Rightarrow \\left(x+\\dfrac {1}{2}\\right)^2+\\dfrac {3}{4} > 0\\,\\forall\\,x) \\\\ \\end{aligned}\\right.$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{2\\}$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span> <br\/><i><span class='basic_green'>Ghi nh\u1edb:<\/span><\/i> Ph\u01b0\u01a1ng tr\u00ecnh $f(x)=0$ (trong \u0111\u00f3, $f(x)$ l\u00e0 m\u1ed9t \u0111a th\u1ee9c) c\u00f3 m\u1ed9t nghi\u1ec7m $x=a$, t\u1ee9c l\u00e0 $f(a)=0$ th\u00ec \u0111a th\u1ee9c $f(x)$ c\u00f3 m\u1ed9t nh\u00e2n t\u1eed l\u00e0 $x-a$. <br\/>Hay<br\/>Ta c\u00f3: $f(x)=0 \\Leftrightarrow (x-a)g(x)=0$. (trong \u0111\u00f3 $g(x)=f(x):(x-a)$)<\/span>","column":2}]}],"id_ques":885},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/9.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $(x^2-4)(x^2-10)=72$ (1).<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{-8;8\\}$","B. $S=\\{4\\}$","C. $S=\\{-4;4\\}$","D. $S=\\varnothing$"],"hint":"\u0110\u1eb7t $t=x^2-4$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1eb7t $x^2-4=t$, \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$.<br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $t$ r\u1ed3i thay v\u00e0o $x^2-4=t$, \u0111\u1ec3 t\u00ecm $x$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110\u1eb7t $x^2-4=t$. Suy ra $x^2-10=t-6$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh:<br\/>$t(t-6)=72\\\\ \\Leftrightarrow t^2-6t=72 \\\\ \\Leftrightarrow t^2-6t-72=0 \\\\ \\Leftrightarrow t^2-12t+6t-72=0 \\\\ \\Leftrightarrow t(t-12)+6(t-12)=0 \\\\ \\Leftrightarrow (t-12)(t+6)=0 \\\\ \\Leftrightarrow \\left [\\begin {aligned}&t=12\\\\&t=-6\\\\ \\end {aligned}\\right.$ <br\/>V\u1edbi $t=12$, ta c\u00f3:<br\/>$x^2-4=12\\\\ \\Leftrightarrow x^2=16\\\\ \\Leftrightarrow x=\\pm 4$<br\/>V\u1edbi $t=-6$, ta c\u00f3: <br\/>$x^2-4=-6\\\\ \\Leftrightarrow x^2=-2\\,\\,\\,\\text {(v\u00f4 nghi\u1ec7m v\u00ec } x^2\\ge 0 \\forall x)$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-4;4\\}$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":886},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/10.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $(x^2+5x)^2+10(x^2+5x)+24=0$ (1).<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{4;6\\}$","B. $S=\\{1;2;3;4\\}$","C. $S=\\{-1;-2;-3;-4\\}$","D. $S=\\varnothing$"],"hint":"\u0110\u1eb7t $t=x^2+5x$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span> <br\/><b>B\u01b0\u1edbc 1:<\/b> \u0110\u1eb7t $x^2+5x=t$ (*)<br\/><b>B\u01b0\u1edbc 2:<\/b> \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$ v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch, r\u1ed3i gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $t$.<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay gi\u00e1 tr\u1ecb $t$ t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o (*) \u1edf b\u01b0\u1edbc 1, gi\u1ea3i t\u00ecm $x$ v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110\u1eb7t $t=x^2+5x$.<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned} & {{t}^{2}}+10t+24=0 \\\\ & \\Leftrightarrow {{t}^{2}}+6t+4t+24=0\\\\ & \\Leftrightarrow t(t+6)+4(t+6)=0 \\\\ & \\Leftrightarrow(t+4)(t+6)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t+4=0 \\\\ & t+6=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=-4 \\\\ & t=-6\\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1edbi $t=-4$, ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{2}}+5x=-4 \\\\ & \\Leftrightarrow {{x}^{2}}+5x+4=0 \\\\ & \\Leftrightarrow {{x}^{2}}+x+4x+4=0 \\\\ & \\Leftrightarrow x(x+1)+4(x+1)=0 \\\\ & \\Leftrightarrow (x+4)(x+1)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-4 \\\\ &x=-1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1edbi $t=-6$, ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{2}}+5x=-6 \\\\ & \\Leftrightarrow {{x}^{2}}+5x+6=0 \\\\ & \\Leftrightarrow {{x}^{2}}+2x+3x+6=0 \\\\ & \\Leftrightarrow x(x+2)+3(x+2)=0 \\\\ & \\Leftrightarrow (x+2)(x+3)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-2 \\\\ & x=-3 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-1;-2;-3;-4\\}$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span> <br\/><b>Nh\u1eadn x\u00e9t:<\/b> B\u00e0i gi\u1ea3i tr\u00ean cho ta m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch \u0111\u00f3 l\u00e0 ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t \u1ea9n ph\u1ee5.<br\/>- Khi ph\u01b0\u01a1ng tr\u00ecnh xu\u1ea5t hi\u1ec7n nh\u1eefng bi\u1ec3u th\u1ee9c ch\u1ee9a ph\u1ea7n \u1ea9n gi\u1ed1ng nhau v\u00e0 l\u1eb7p l\u1ea1i ta c\u00f3 th\u1ec3 th\u1ef1c hi\u1ec7n \u0111\u1eb7t bi\u1ec3u th\u1ee9c l\u1eb7p l\u1ea1i l\u00e0 $t$ \u0111\u1ec3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$.<br\/>- Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$, t\u00ecm $t$ r\u1ed3i t\u00ecm $x$.<\/span>","column":2}]}],"id_ques":887},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/9.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $(2x^2+x-6)^2+3(2x^2+x-3)-9=0$ (1).<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{ 1;\\dfrac {-3}{2}\\right\\}$","B. $S=\\left\\{-2;\\dfrac {3}{2}\\right\\}$","C. $S=\\left\\{-2;\\dfrac {-3}{2}; 1;\\dfrac {3}{2}\\right\\}$","D. $S=\\left\\{0;\\dfrac {-3}{2}\\right\\}$"],"hint":"\u0110\u1eb7t $t=2x^2+x-6$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u1eb7t $2x^2+x-6=t$, \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$.<br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $t$ r\u1ed3i thay v\u00e0o $t=2x^2+x-6$, \u0111\u1ec3 t\u00ecm $x$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110\u1eb7t $2x^2+x-6=t$ th\u00ec $2x^2+x-3=t+3$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh:<br\/>$t^2+3(t+3)-9=0\\\\ \\Leftrightarrow t^2+3t=0 \\\\ \\Leftrightarrow t(t+3)=0 \\\\ \\Leftrightarrow \\left [\\begin {aligned}&t=0\\\\&t=-3\\\\ \\end {aligned}\\right.$ <br\/>V\u1edbi $t=0$, ta c\u00f3:<br\/>$2x^2+x-6=0\\\\ \\Leftrightarrow (x+2)(2x-3)=0\\\\ \\Leftrightarrow \\left [ \\begin {aligned} & x=-2\\\\ & x=\\dfrac {3}{2}\\\\ \\end {aligned}\\right.$<br\/>V\u1edbi $t=-3$, ta c\u00f3: <br\/>$2x^2+x-6=-3\\\\ \\Leftrightarrow 2x^2+x-3=0\\\\ \\Leftrightarrow (x-1)(2x+3)=0 \\\\ \\Leftrightarrow \\left[\\begin{aligned} &x-1=0\\\\ &2x+3=0\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin{aligned} &x=1\\\\ &x=\\dfrac {-3}{2}\\\\ \\end{aligned}\\right.$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\left\\{-2;\\dfrac {-3}{2}; 1;\\dfrac {3}{2}\\right\\}$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span> <\/span>","column":2}]}],"id_ques":888},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"fill_the_blank_random","correct":[[["-4"],["1"]]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/2.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x(x+1)(x+2)(x+3)=24$ .<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$","hint":"Nh\u00f3m nh\u00e2n t\u1eed th\u1ee9 nh\u1ea5t v\u1edbi nh\u00e2n t\u1eed th\u1ee9 4, nh\u00e2n t\u1eed th\u1ee9 2 v\u1edbi nh\u00e2n t\u1eed th\u1ee9 3. Th\u1ef1c hi\u1ec7n ph\u00e9p nh\u00e2n r\u1ed3i \u0111\u1eb7t \u1ea9n.","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b>B\u01b0\u1edbc 1:<\/b> Nh\u00e2n $x$ v\u1edbi $x+3$, v\u00e0 $x+1$ v\u1edbi $x+2$<br\/> <b>B\u01b0\u1edbc 2:<\/b> \u0110\u1eb7t $x^2+3x =t$.<br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$.<br\/><b>B\u01b0\u1edbc 4:<\/b> Thay gi\u00e1 tr\u1ecb c\u1ee7a $t$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u1edf b\u01b0\u1edbc 2 \u0111\u1ec3 t\u00ecm $x$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & x(x+1)(x+2)(x+3)=24 \\\\ & \\Leftrightarrow \\left[ x(x+3) \\right]\\left[ (x+1)(x+2) \\right]=24 \\\\ & \\Leftrightarrow ({{x}^{2}}+3x)({{x}^{2}}+3x+2)=24 \\\\ \\end{aligned}$<br\/>\u0110\u1eb7t $x^2+3x=t$. Ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned} & t\\left( t+2 \\right)=24\\\\ & \\Leftrightarrow {{t}^{2}}+2t-24=0 \\\\ & \\Leftrightarrow (t+1)^2-25=0\\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t+1=-5 \\\\ & t+1=5 \\\\ \\end{aligned} \\right.\\\\& \\Leftrightarrow \\left[ \\begin{aligned} & t=-6 \\\\ & t=4 \\\\ \\end{aligned} \\right.\\\\ \\end{aligned}$<br\/>V\u1edbi $t=-6$, ta c\u00f3:<br\/>$\\begin{align} & {{x}^{2}}+3x=-6 \\\\ & \\Leftrightarrow {{x}^{2}}+3x+6=0 \\\\ & \\Leftrightarrow \\left(x+\\dfrac{3}{2}\\right)^{2}=\\dfrac{-15}{4} \\,\\,\\,\\text{(lo\u1ea1i)} \\\\ \\end{align}$<br\/>V\u1edbi $t=4$, ta c\u00f3:<br\/>$\\begin{align} & {{x}^{2}}+3x=4 \\\\ & \\Leftrightarrow {{x}^{2}}+3x-4=0 \\\\ & \\Leftrightarrow x^2-x+4x-4=0\\\\&\\Leftrightarrow x(x-1)+4(x-1)=0 \\\\ &\\Leftrightarrow (x-1)(x+4)=0\\\\ &\\Leftrightarrow \\left[ \\begin{aligned} & x-1=0 \\\\ & x+4=0 \\\\ \\end{aligned} \\right.\\\\ &\\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & x=-4 \\\\ \\end{aligned} \\right.\\\\ \\end{align}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-4;1\\}$.<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-4;\\,1$<\/span> <br\/><b>Nh\u1eadn x\u00e9t:<\/b> Ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean l\u00e0 m\u1ed9t d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0a \u0111\u01b0\u1ee3c v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch \u0111\u1ec3 gi\u1ea3i.<br\/> B\u00e0i to\u00e1n \u0111\u01b0a ra m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng: $(x+a)(x+b)(x+c)(x+d)=m$.<br\/> Trong \u0111\u00f3 $a,\\,b,\\,c,\\,d\\in \\mathbb Z , m \\in \\mathbb R; a+c=b+d$. \u0110\u00f3 l\u00e0:<br\/>Nh\u00f3m nh\u00e2n t\u1eed $x+a$ v\u1edbi $x+c$ v\u00e0 $x+b$ v\u1edbi $x+d$, th\u1ef1c hi\u1ec7n ph\u00e9p nh\u00e2n r\u1ed3i ph\u00e1t hi\u1ec7n \u1ea9n ph\u1ee5. <\/span>"}]}],"id_ques":889},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. {$\\dfrac{-2}{3}; \\dfrac{-5}{3}$}","B. {$\\dfrac{2}{3}; \\dfrac{-5}{3}$}","C. {$\\dfrac{-2}{3}; \\dfrac{5}{3}$}"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv3/img\/12.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $(6x+7)^2(3x+4)(x+1)=6.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=?$","hint":"Nh\u00e2n $3x+4$ v\u1edbi $2$ v\u00e0 $x+1$ v\u1edbi $6$. \u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh m\u1edbi l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho, v\u1ebf ph\u1ea3i nh\u00e2n v\u1edbi $12$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 xu\u1ea5t hi\u1ec7n bi\u1ec3u th\u1ee9c \u0111\u1eb7t \u1ea9n ph\u1ee5<br\/><b>B\u01b0\u1edbc 2:<\/b> \u0110\u1eb7t \u1ea9n ph\u1ee5 $t$, gi\u1ea3i t\u00ecm $t$.<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay gi\u00e1 tr\u1ecb $t$ t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u1edf b\u01b0\u1edbc 2, \u0111\u1ec3 t\u00ecm $x$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/>$\\begin{align} & {{(6x+7)}^{2}}(3x+4)(x+1)=6 \\\\ & \\Leftrightarrow {{(6x+7)}^{2}}(6x+8)(6x+6)=72 \\\\\\end{align}$<br\/>\u0110\u1eb7t $6x+7=t$, suy ra $6x+8=t+1;\\,6x+6=t-1$.<br\/> Ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u tr\u1edf th\u00e0nh:<br\/>$t^2(t+1)(t-1)=72\\\\ \\Leftrightarrow t^2(t^2-1)=72\\\\ \\Leftrightarrow t^4-t^2-72=0 \\\\ \\Leftrightarrow t^4-9t^2+8t^2-72=0\\\\ \\Leftrightarrow t^2(t^2-9)+8(t^2-9)=0\\\\ \\Leftrightarrow (t^2-9)(t^2+8)=0\\\\ \\Leftrightarrow \\left [\\begin{aligned}&t^2=9\\\\ &t^2=-8\\,\\,\\,\\,\\text {(lo\u1ea1i)}\\\\ \\end{aligned}\\right. \\\\ \\Leftrightarrow t=\\pm 3$<br\/>V\u1edbi $t=3$, ta c\u00f3:<br\/>$6x+7=3\\\\ \\Leftrightarrow 6x=-4 \\\\ \\Leftrightarrow x=\\dfrac {-2} {3}$<br\/>V\u1edbi $t=-3$, ta c\u00f3:<br\/>$6x+7=-3\\\\ \\Leftrightarrow 6x=-10\\\\ \\Leftrightarrow x=\\dfrac {-5}{3}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\left\\{\\dfrac {-2} {3};\\dfrac {-5}{3}\\right\\}$<br\/><b>Nh\u1eadn x\u00e9t:<\/b> T\u1eeb b\u00e0i to\u00e1n tr\u00ean ta nh\u1eadn th\u1ea5y r\u1eb1ng: M\u1ed9t s\u1ed1 ph\u01b0\u01a1ng tr\u00ecnh c\u1ea7n th\u1ef1c hi\u1ec7n m\u1ed9t s\u1ed1 ph\u00e9p bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u01a1 b\u1ea3n \u0111\u1ec3 c\u00f3 th\u1ec3 th\u1ef1c hi\u1ec7n ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t \u1ea9n ph\u1ee5 \u0111\u1ec3 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.<\/span>"}]}],"id_ques":890}],"lesson":{"save":0,"level":3}}

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Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên + 5 điểm
Trong khoảng 1 phút -> 2 phút + 4 điểm
Trong khoảng 2 phút -> 3 phút + 3 điểm
Trong khoảng 3 phút -> 4 phút + 2 điểm
Trong khoảng 4 phút -> 5 phút + 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

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