{"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}}