{"segment":[{"time":24,"part":[{"time":3,"title":"N\u1ed1i m\u1ed7i ph\u01b0\u01a1ng tr\u00ecnh c\u1ed9t tr\u00e1i v\u1edbi \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh t\u01b0\u01a1ng \u1ee9ng \u1edf c\u1ed9t ph\u1ea3i. ","title_trans":"","audio":"","temp":"matching","correct":[["2","1","4","3"]],"list":[{"point":10,"image":"","left":["$\\dfrac{x+1}{{{x}^{3}}+1}=\\dfrac{1}{{{x}^{2}}-x-2}$ ","$\\dfrac{1}{x}+5=\\dfrac{8}{{{x}^{2}}-4x+3}$","$\\dfrac{x+1}{{{x}^{2}}-5x+6}=\\dfrac{1}{x-2}+\\dfrac{2}{2x-6}$ ","$\\dfrac{1}{{{x}^{2}}+1}-\\dfrac{1}{{{x}^{2}}+2x-3}=\\dfrac{x+1}{x-1}$"],"right":["$\\left\\{ \\begin{align}& x\\ne 0 \\\\ & x\\ne 1 \\\\ & x\\ne 3 \\\\ \\end{align} \\right.$","$\\left\\{ \\begin{align} & x\\ne -1 \\\\ & x\\ne 2 \\\\ \\end{align} \\right.$","$\\left\\{ \\begin{align} & x\\ne 1 \\\\ & x\\ne -3 \\\\ \\end{align} \\right.$","$\\left\\{ \\begin{align}& x\\ne 2 \\\\ & x\\ne 3 \\\\ \\end{align} \\right.$"],"top":100,"hint":"","explain":"<span class='basic_left'>+) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\dfrac{x+1}{{{x}^{3}}+1}=\\dfrac{1}{{{x}^{2}}-x-2}$ l\u00e0: <br\/>$\\left\\{ \\begin{aligned}& {{x}^{3}}+1\\ne 0 \\\\ & {{x}^{2}}-x-2\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & (x+1)({{x}^{2}}+x+1)\\ne 0 \\\\ & (x+1)(x-2)\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne -1 \\\\ & x\\ne 2 \\\\ \\end{aligned} \\right.$<br\/>+) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\dfrac{1}{x}+5=\\dfrac{8}{{{x}^{2}}-4x+3}$ l\u00e0:<br\/>$\\left\\{ \\begin{aligned}& x\\ne 0 \\\\ & {{x}^{2}}-4x+3\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & (x-1)(x-3)\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x\\ne 1 \\\\ & x\\ne 3 \\\\ \\end{aligned} \\right.$<br\/>+) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\dfrac{x+1}{{{x}^{2}}-5x+6}=\\dfrac{1}{x-2}+\\dfrac{2}{2x-6}$ l\u00e0: <br\/>$\\left\\{ \\begin{aligned} & {{x}^{2}}-5x+6\\ne 0 \\\\ & x-2\\ne 0 \\\\ & 2x-6\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned}& (x-2)(x-3)\\ne 0 \\\\ & x\\ne 2 \\\\ & 2x\\ne 6 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne 2 \\\\ & x\\ne 3 \\\\ \\end{aligned} \\right.$<br\/>+) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\dfrac{1}{{{x}^{2}}+1}-\\dfrac{1}{{{x}^{2}}+2x-3}=\\dfrac{x+1}{x-1}$ l\u00e0:<br\/>$\\left\\{ \\begin{aligned}& {{x}^{2}}+1\\ne 0 \\\\ & {{x}^{2}}+2x-3\\ne 0 \\\\ & x-1\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned}& {{x}^{2}}\\ne -1\\,\\,\\forall x \\\\ & (x-1)(x+3)\\ne 0 \\\\ & x\\ne 1 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne 1 \\\\ & x\\ne -3 \\\\ \\end{aligned} \\right.$<\/span>"}]}],"id_ques":931},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv3/img\/10.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac {2}{x-2}+\\dfrac {3}{x-3}=\\dfrac {4}{x-4}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{6+2\\sqrt{3}\\}$","B. $S=\\{6-2\\sqrt{3};6+2\\sqrt{3}\\}$","C. $S=\\mathbb R$","D. $S=\\varnothing$"],"hint":"Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c, kh\u1eed m\u1eabu \u0111\u01b0a v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch. <br\/>S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $f^2(x)-b=0\\,\\,(b>0)$. ","explain":" <span class='basic_left'><br\/><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh.<br\/><b>B\u01b0\u1edbc 2:<\/b> Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu thu \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh m\u1edbi.<br\/><b>B\u01b0\u1edbc 3:<\/b> Ph\u00e2n t\u00edch $x^2-12x+24=(x-6)^2-12$<br\/><b>B\u01b0\u1edbc 4:<\/b> \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch r\u1ed3i gi\u1ea3i v\u00e0 k\u1ebft lu\u1eadn nghi\u1ec7m.<\/span> <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2; x\\ne 3$ v\u00e0 $x\\ne 4$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{2}{x-2}+\\dfrac{3}{x-3}=\\dfrac{4}{x-4} \\\\ & \\Leftrightarrow \\dfrac{2(x-3)(x-4)}{(x-2)(x-3)(x-4)}+\\dfrac{3(x-2)(x-4)}{(x-2)(x-3)(x-4)}=\\dfrac{4(x-2)(x-3)}{(x-2)(x-3)(x-4)} \\\\ & \\Rightarrow 2(x-3)(x-4)+3(x-2)(x-4)=4(x-2)(x-3) \\\\ & \\Leftrightarrow 2({{x}^{2}}-7x+12)+3({{x}^{2}}-6x+8)-4({{x}^{2}}-5x+6)=0 \\\\ & \\Leftrightarrow {{x}^{2}}-12x+24=0 \\\\ & \\Leftrightarrow {{x}^{2}}-2x.6+36-12=0 \\\\ & \\Leftrightarrow {{(x-6)}^{2}}=12 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-6=2\\sqrt{3} \\\\ & x-6=-2\\sqrt{3} \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=6+2\\sqrt{3}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=6-2\\sqrt{3}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{6-2\\sqrt{3};6+2\\sqrt{3}\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":932},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv3/img\/2.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x+1}{{{x}^{2}}+x+1}-\\dfrac{x-1}{{{x}^{2}}-x+1}=\\dfrac{3}{{{x}^{4}}+{{x}^{2}}+1}$.<br\/> T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{1\\}$","B. $S=\\{-1;0\\}$","C. $S=\\mathbb R$","D. $S=\\varnothing$"],"hint":"Nh\u00e2n m\u1eabu th\u1ee9c c\u1ee7a ph\u00e2n th\u1ee9c th\u1ee9 nh\u1ea5t v\u00e0 m\u1eabu th\u1ee9c c\u1ee7a ph\u00e2n th\u1ee9c th\u1ee9 hai, s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$ \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh m\u1eabu th\u1ee9c chung.","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh.<br\/><b>B\u01b0\u1edbc 2:<\/b> Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c: M\u1eabu th\u1ee9c chung l\u00e0 $(x^2-x+1)(x^2+x+1)=x^4+x^2+1$<br\/><b>B\u01b0\u1edbc 3:<\/b> R\u00fat g\u1ecdn v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.<br\/><b>B\u01b0\u1edbc 4:<\/b> Lo\u1ea1i nghi\u1ec7m (n\u1ebfu c\u00f3) v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $\\left\\{ \\begin{aligned} & {{x}^{2}}+x+1\\ne 0 \\\\ & {{x}^{2}}-x+1\\ne 0 \\\\ & {{x}^{4}}+{{x}^{2}}+1\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned}& {{\\left( x+\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{3}{4}\\ne 0 \\\\ & {{\\left( x-\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{3}{4}\\ne 0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\forall x\\in \\mathbb{R} \\\\ & {{\\left( {{x}^{2}}+\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{3}{4}\\ne 0 \\\\ \\end{aligned} \\right.$<br\/>Ta c\u00f3:<br\/>$\\begin{align}& \\dfrac{x+1}{{{x}^{2}}+x+1}-\\dfrac{x-1}{{{x}^{2}}-x+1}=\\dfrac{3}{{{x}^{4}}+{{x}^{2}}+1} \\\\ & \\Leftrightarrow \\dfrac{(x+1)({{x}^{2}}-x+1)}{({{x}^{2}}+x+1)({{x}^{2}}-x+1)}-\\dfrac{(x-1)({{x}^{2}}+x+1)}{({{x}^{2}}+x+1)({{x}^{2}}-x+1)}=\\dfrac{3}{({{x}^{2}}+x+1)({{x}^{2}}-x+1)} \\\\ & \\Rightarrow (x+1)({{x}^{2}}-x+1)-(x-1)({{x}^{2}}+x+1)=3 \\\\ & \\Leftrightarrow {{x}^{3}}+1-({{x}^{3}}-1)=3 \\\\ & \\Leftrightarrow {{x}^{3}}+1-{{x}^{3}}+1=3 \\\\ & \\Leftrightarrow 2=3\\,\\,\\,\\,\\,\\,\\,\\,(\\text{v\u00f4 l\u00fd}) \\\\ \\end{align}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":933},{"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/bai20/lv3/img\/11.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $2\\left( {{x}^{2}}+\\dfrac{1}{{{x}^{2}}} \\right)-\\left( x+\\dfrac{1}{x} \\right)=6$ c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{2;-1\\}$. <b> \u0110\u00fang<\/b> hay <b> Sai <\/b>? ","select":["\u0110\u00fang","Sai"],"hint":"Bi\u1ebfn \u0111\u1ed5i v\u1ebf tr\u00e1i, r\u1ed3i \u0111\u1eb7t $x+\\dfrac{1}{x}=t$","explain":" <span class='basic_left'><span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 0$<br\/>Ta c\u00f3:<br\/>$\\begin{align} & 2\\left( {{x}^{2}}+\\dfrac{1}{{{x}^{2}}} \\right)-\\left( x+\\dfrac{1}{x} \\right)=6 \\\\ & \\Leftrightarrow 2\\left( {{x}^{2}}+2.x.\\dfrac{1}{x}+\\dfrac{1}{{{x}^{2}}}-2 \\right)-\\left( x+\\dfrac{1}{x} \\right)=6 \\\\ & \\Leftrightarrow 2{{\\left( x+\\dfrac{1}{x} \\right)}^{2}}-\\left( x+\\dfrac{1}{x} \\right)-4-6=0 \\\\ & \\Leftrightarrow 2{{\\left( x+\\dfrac{1}{x} \\right)}^{2}}-\\left( x+\\dfrac{1}{x} \\right)-10=0 \\\\ \\end{align}$. <br\/>\u0110\u1eb7t $x+\\dfrac{1}{x}=t$. Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned}& 2{{t}^{2}}-t-10=0 \\\\ & \\Leftrightarrow 2{{t}^{2}}-5t+4t-10=0 \\\\ & \\Leftrightarrow t(2t-5)+2\\left( 2t-5 \\right)=0 \\\\ & \\Leftrightarrow \\left( 2t-5 \\right)\\left( t+2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2t-5=0 \\\\ & t+2=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=\\dfrac{5}{2} \\\\ & t=-2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1edbi $t=-2$, ta c\u00f3:<br\/>$x+\\dfrac{1}{x}=-2\\\\ \\Leftrightarrow x^2+2x+1=0\\\\ \\Leftrightarrow (x+1)^2=0\\\\ \\Leftrightarrow x=-1\\,\\,\\text{(th\u1ecfa m\u00e3n)}$<br\/>V\u1edbi $t=\\dfrac{5}{2}$, ta c\u00f3:<br\/>$x+\\dfrac{1}{x}=\\dfrac{5}{2}\\\\ \\Leftrightarrow 2x^2-5x+2=0\\\\ \\Leftrightarrow 2x^2-4x-x+2=0\\\\ \\Leftrightarrow 2x(x-2)-(x-2)=0\\\\ \\Leftrightarrow (2x-1)(x-2)=0\\\\ \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{1}{2}\\,\\,\\text{(th\u1ecfa m\u00e3n)} \\\\ & x=2 \\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{aligned} \\right. $<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{-1;\\dfrac{1}{2};2\\right\\}$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 sai<\/span><\/span>","column":2}]}],"id_ques":934},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"fill_the_blank_random","correct":[[["-2"],["0"],["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv3/img\/12.jpg' \/><\/center>Cho ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x+2}{x-m}=\\dfrac{x+1}{x-1}$ ($m$ l\u00e0 tham s\u1ed1). T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\,m\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ v\u00e0 $m\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","hint":"Ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu c\u00f3 nghi\u1ec7m khi t\u1ea5t c\u1ea3 gi\u00e1 tr\u1ecb $x$ t\u00ecm \u0111\u01b0\u1ee3c ph\u1ea3i th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh.","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne m; x\\ne 1$<br\/>Ta c\u00f3:<br\/>$\\begin{align}& \\dfrac{x+2}{x-m}=\\dfrac{x+1}{x-1} \\,\\,\\,(*)\\\\ & \\Leftrightarrow \\dfrac{(x+2)(x-1)}{(x-1)(x-m)}=\\dfrac{(x+1)(x-m)}{(x-1)(x-m)} \\\\ & \\Rightarrow (x+2)(x-1)-(x+1)(x-m)=0 \\\\ & \\Leftrightarrow {{x}^{2}}+x-2-({{x}^{2}}+x-mx-m)=0 \\\\ & \\Leftrightarrow {{x}^{2}}+x-2-{{x}^{2}}-x+mx+m=0 \\\\ & \\Leftrightarrow mx+m-2=0 \\,\\,\\,(1)\\\\ \\end{align}$<br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 nghi\u1ec7m th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 nghi\u1ec7m kh\u00e1c $1$ v\u00e0 kh\u00e1c $m$<br\/><b> $+ $<\/b> Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 nghi\u1ec7m kh\u00e1c $1$ hay:<br\/> $\\left\\{ \\begin{aligned} & m.1+m-2\\ne 0 \\\\ & m\\ne 0 \\\\ \\end{aligned} \\right. \\\\ \\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 1 \\\\ & m\\ne 0 \\\\ \\end{aligned} \\right.$<br\/><b> $+$<\/b> Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 nghi\u1ec7m kh\u00e1c $m$ hay:<br\/> $\\left\\{ \\begin{aligned} & m.m+m-2\\ne 0 \\\\ & m\\ne 0 \\\\ \\end{aligned} \\right. \\\\ \\Leftrightarrow \\left\\{ \\begin{aligned} & m^2+m-2\\ne 0 \\\\ & m\\ne 0 \\\\ \\end{aligned} \\right.\\\\\\Leftrightarrow \\left\\{ \\begin{aligned} & (m-1)(m+2) \\ne 0 \\\\ & m\\ne 0 \\\\ \\end{aligned} \\right.\\\\ \\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 1 \\\\ & m\\ne -2\\\\ & m\\ne 0 \\\\ \\end{aligned} \\right.$<br\/>V\u1eady \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u00e2u c\u00f3 nghi\u1ec7m th\u00ec $m\\ne 0;\\,m\\ne -2$ v\u00e0 $m\\ne 1$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2;0$ v\u00e0 $1$<\/span><br\/><i><span class='basic_green'>Ghi nh\u1edb:<\/span><\/i> <br\/>Ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu c\u00f3 nghi\u1ec7m khi gi\u00e1 tr\u1ecb $x$ t\u00ecm \u0111\u01b0\u1ee3c ph\u1ea3i th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh.<br\/> Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n d\u1ea1ng $ax+b=0$ c\u00f3:<br\/>+) $1$ nghi\u1ec7m duy nh\u1ea5t khi $a\\ne 0$<br\/>+) V\u00f4 s\u1ed1 nghi\u1ec7m khi $a=0$ v\u00e0 $b=0$<br\/>+) V\u00f4 nghi\u1ec7m khi $a=0$ v\u00e0 $b\\ne 0$<\/span>"}]}],"id_ques":935},{"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/bai20/lv3/img\/11.jpg' \/><\/center>Cho ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x+m}{x+1}+\\dfrac{x-2}{x}=2$ ($m$ l\u00e0 tham s\u1ed1). T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m.","select":["A. $m=3$","B. $m\\ne 3$","C. $m\\in \\{1;3\\}$","D. $m=1$"],"hint":"Quy \u0111\u1ed3ng kh\u1eed m\u1eabu r\u1ed3i bi\u1ec7n lu\u1eadn s\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh theo $m$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu c\u00f3 nghi\u1ec7m khi t\u1ea5t c\u1ea3 gi\u00e1 tr\u1ecb $x$ t\u00ecm \u0111\u01b0\u1ee3c ph\u1ea3i th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh.","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne -1; x\\ne 0$<br\/>Ta c\u00f3:<br\/>$\\begin{align}& \\dfrac{x+m}{x+1}+\\dfrac{x-2}{x}=2 \\,\\,\\,(*)\\\\ & \\Leftrightarrow \\dfrac{x(x+m)}{x(x+1)}+\\dfrac{(x-2)(x+1)}{x(x+1)}=\\dfrac{2x(x+1)}{x(x+1)} \\\\ & \\Rightarrow x(x+m)+({{x}^{2}}-x-2)=2x(x+1) \\\\ & \\Leftrightarrow {{x}^{2}}+mx+{{x}^{2}}-x-2-2{{x}^{2}}-2x=0 \\\\ & \\Leftrightarrow (m-3)x=2 \\,\\,\\,\\,\\,\\,\\,\\,(1)\\\\ \\end{align}$<br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (*) v\u00f4 nghi\u1ec7m th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 nghi\u1ec7m $x=-1$ ho\u1eb7c $x=0$ ho\u1eb7c v\u00f4 nghi\u1ec7m.<br\/><b>Tr\u01b0\u1eddng h\u1ee3p 1:<\/b> Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 m\u1ed9t nghi\u1ec7m $x=-1$.<br\/>Khi \u0111\u00f3, ta c\u00f3: $(m-3)(-1)=2\\Leftrightarrow m=1$<br\/><b>Tr\u01b0\u1eddng h\u1ee3p 2:<\/b> Ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 m\u1ed9t nghi\u1ec7m $x=0$.<br\/>Khi \u0111\u00f3, ta c\u00f3: $(m-3).0=2 \\Leftrightarrow 0=2\\,\\,\\, \\text {(v\u00f4 l\u00fd)}$<br\/><b>Tr\u01b0\u1eddng h\u1ee3p 3:<\/b> Ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m: $m=3$<br\/>V\u1eady \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u v\u00f4 nghi\u1ec7m $m\\in \\{1;3\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><br\/><i><span class='basic_green'>Ghi nh\u1edb:<\/span><\/i> Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n d\u1ea1ng $ax+b=0$ c\u00f3:<br\/>+) $1$ nghi\u1ec7m duy nh\u1ea5t khi $a\\ne 0$<br\/>+) V\u00f4 s\u1ed1 nghi\u1ec7m khi $a=0$ v\u00e0 $b=0$<br\/>+) V\u00f4 nghi\u1ec7m khi $a=0$ v\u00e0 $b\\ne 0$<\/span>","column":2}]}],"id_ques":936},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 d\u01b0\u1edbi d\u1ea1ng s\u1ed1 ph\u00e2n s\u1ed1","temp":"fill_the_blank","correct":[[["-7"],["4"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":"frac","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv3/img\/9.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x+4}{2{{x}^{2}}-5x+2}+\\dfrac{x+1}{2{{x}^{2}}-7x+3}=\\dfrac{-x}{{{x}^{2}}-5x+6}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>}","hint":"Nh\u1ea9m nghi\u1ec7m nguy\u00ean c\u1ee7a m\u1ed7i m\u1eabu th\u1ee9c r\u1ed3i ph\u00e2n t\u00edch m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed \u0111\u1ec3 x\u00e1c \u0111\u1ecbnh m\u1eabu th\u1ee9c chung v\u00e0 quy \u0111\u1ed3ng m\u1eabu th\u1ee9c.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Ph\u00e2n t\u00edch m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed:<br\/>$2{{x}^{2}}-5x+2=(x-2)(2x-1)$<br\/>$2{{x}^{2}}-7x+3=(2x-1)(x-3)$<br\/>${{x}^{2}}-5x+6=(x-2)(x-3)$<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh v\u00e0 m\u1eabu th\u1ee9c chung<br\/><b>B\u01b0\u1edbc 3:<\/b> Quy \u0111\u1ed3ng, kh\u1eed m\u1eabu v\u00e0 r\u00fat g\u1ecdn ph\u01b0\u01a1ng tr\u00ecnh<br\/><b>B\u01b0\u1edbc 4:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 k\u1ebft lu\u1eadn nghi\u1ec7m<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/>$2{{x}^{2}}-5x+2=2{{x}^{2}}-4x-x+2=2x(x-2)-(x-2)=(x-2)(2x-1)$<br\/>$2{{x}^{2}}-7x+3=2{{x}^{2}}-6x-x+3=2x(x-3)-(x-3)=(2x-1)(x-3)$<br\/>${{x}^{2}}-5x+6={{x}^{2}}-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)$<br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2; x\\ne 3$ v\u00e0 $x\\ne \\dfrac {1}{2}$<br\/>$\\begin{aligned} & \\dfrac{x+4}{2{{x}^{2}}-5x+2}+\\dfrac{x+1}{2{{x}^{2}}-7x+3}=\\dfrac{-x}{{{x}^{2}}-5x+6} \\\\ & \\Leftrightarrow \\dfrac{x+4}{(2x-1)(x-2)}+\\dfrac{x+1}{(2x-1)(x-3)}=\\dfrac{-x}{(x-2)(x-3)} \\\\ & \\Leftrightarrow \\dfrac{(x+4)(x-3)}{(2x-1)(x-2)(x-3)}+\\dfrac{(x+1)(x-2)}{(2x-1)(x-2)(x-3)}=\\dfrac{-x(2x-1)}{(2x-1)(x-2)(x-3)} \\\\ & \\Rightarrow (x+4)(x-3)+(x+1)(x-2)=-2{{x}^{2}}+x \\\\ & \\Leftrightarrow {{x}^{2}}+x-12+{{x}^{2}}-x-2+2{{x}^{2}}-x=0 \\\\ & \\Leftrightarrow 4{{x}^{2}}-x-14=0 \\\\ & \\Leftrightarrow 4{{x}^{2}}-8x+7x-14=0 \\\\ & \\Leftrightarrow 4x(x-2)+7(x-2)=0 \\\\ & \\Leftrightarrow (4x+7)(x-2)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 4x+7=0 \\\\ & x-2=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{-7}{4}\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{\\dfrac{-7}{4}\\}$<br\/><i><span class='basic_green'>Ghi nh\u1edb: <\/span><\/i>Khi ph\u00e2n t\u00edch \u0111a th\u1ee9c $f(x)$ th\u00e0nh nh\u00e2n t\u1eed:<br\/>N\u1ebfu x\u00e1c \u0111\u1ecbnh \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 $a$ sao cho $f(a)=0$, ta lu\u00f4n ph\u00e2n t\u00edch \u0111\u01b0\u1ee3c $f(x)$ v\u1ec1 d\u1ea1ng $(x-a)g(x)$ trong \u0111\u00f3 $g(x) $ l\u00e0 \u0111a th\u1ee9c \u1ea9n $x$ b\u1eadc nh\u1ecf h\u01a1n $f(x)$<\/span>"}]}],"id_ques":937},{"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/bai20/lv3/img\/10.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x}{{{x}^{2}}-5x+3}-\\dfrac{2x}{{{x}^{2}}+x+3}=-1.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{2;3\\}$","B. $S=\\varnothing$","C. $S=\\{0;1\\}$","D. $S=\\mathbb R$"],"hint":"Chia c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a hai ph\u00e2n th\u1ee9c cho $x$ r\u1ed3i \u0111\u1eb7t \u1ea9n v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":" <span class='basic_left'>Nh\u1eadn x\u00e9t: $x=0$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/>Chia c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a hai ph\u00e2n th\u1ee9c c\u1ee7a v\u1ebf tr\u00e1i cho $x$ ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{x}{{{x}^{2}}-5x+3}-\\dfrac{2x}{{{x}^{2}}+x+3}=-1 \\\\ & \\Leftrightarrow \\dfrac{1}{x-5+\\dfrac{3}{x}}-\\dfrac{2}{x+1+\\dfrac{3}{x}}=-1 \\\\ & \\Leftrightarrow \\dfrac{1}{x+\\dfrac{3}{x}-5}-\\dfrac{2}{x+\\dfrac{3}{x}+1}=-1\\,\\,\\,\\,\\,(*) \\\\ \\end{aligned}$<br\/>\u0110\u1eb7t $x+\\dfrac {3}{x} =t$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (*) tr\u1edf th\u00e0nh<br\/>$\\begin{aligned} & \\dfrac{1}{t-5}-\\dfrac{2}{t+1}=-1 \\\\ & \\Leftrightarrow \\dfrac{t+1}{(t-5)(t+1)}-\\dfrac{2(t-5)}{(t-5)(t+1)}=\\dfrac{-(t-5)(t+1)}{(t-5)(t+1)} \\\\ & \\Rightarrow t+1-2t+10=-({{t}^{2}}-4t-5) \\\\ & \\Leftrightarrow {{t}^{2}}-4t-5-t+11=0 \\\\ & \\Leftrightarrow {{t}^{2}}-5t+6=0\\\\ &\\Leftrightarrow t^2-2t-3t+6=0\\\\ &\\Leftrightarrow t(t-2)-3(t-2)=0 \\\\ & \\Leftrightarrow (t-2)(t-3)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=2 \\\\ & t=3\\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1edbi $t=2$, ta c\u00f3:<br\/>$\\begin{align} & x+\\dfrac{3}{x}=2 \\\\ & \\Rightarrow {{x}^{2}}-2x+3=0 \\\\ & \\Leftrightarrow {{(x-1)}^{2}}+2=0\\,\\,\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m}) \\\\ \\end{align}$<br\/>V\u1edbi $t=3$, ta c\u00f3:<br\/>$\\begin{aligned} & x+\\dfrac{3}{x}=3 \\\\ & \\Rightarrow {{x}^{2}}-3x+3=0 \\\\ & \\Leftrightarrow {{x}^{2}}-2.x.\\dfrac{3}{2}+\\dfrac{9}{4}+\\dfrac{3}{4}=0 \\\\ & \\Leftrightarrow {{\\left( x-\\dfrac{3}{2} \\right)}^{2}}+\\dfrac{3}{4}=0\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m}) \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho v\u00f4 nghi\u1ec7m<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":938},{"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","5"],["5","-1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv3/img\/13.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{{{(x-2)}^{2}}}{{{x}^{2}}-4x+5}-\\dfrac{{{x}^{2}}-4x+3}{{{(x-2)}^{2}}}=\\dfrac{1}{90}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_;_input_}","hint":"Ph\u00e2n t\u00edch c\u00e1c bi\u1ec3u th\u1ee9c theo h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(a\\pm b)^2$ r\u1ed3i \u0111\u1eb7t $x^2+4x+4=t$. ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2$<br\/>Ta c\u00f3:<br\/> $\\dfrac{{{(x-2)}^{2}}}{{{x}^{2}}-4x+5}-\\dfrac{{{x}^{2}}-4x+3}{{{(x-2)}^{2}}}=\\dfrac{1}{90} \\\\ \\Leftrightarrow \\dfrac{{{x}^{2}}-4x+4}{{{x}^{2}}-4x+5}-\\dfrac{{{x}^{2}}-4x+3}{{{x}^{2}}-4x+4}=\\dfrac{1}{90}$<br\/>\u0110\u1eb7t $x^2-4x+4=t\\,\\,\\,\\,(t\\ge 0)$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned}& \\dfrac{t}{t+1}-\\dfrac{t-1}{t}=\\dfrac{1}{90} \\\\ & \\Leftrightarrow \\dfrac{90{{t}^{2}}}{90t(t+1)}-\\dfrac{90(t-1)(t+1)}{90t(t+1)}=\\dfrac{t(t+1)}{90t(t+1)} \\\\ & \\Rightarrow 90{{t}^{2}}-90({{t}^{2}}-1)=t(t+1) \\\\ & \\Leftrightarrow 90=t(t+1) \\\\ & \\Leftrightarrow {{t}^{2}}+t-90=0 \\\\ & \\Leftrightarrow {{t}^{2}}+10t-9t-90=0 \\\\ & \\Leftrightarrow (t+10)(t-9)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=-10\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ & t=9\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1edbi $t=9$ ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{2}}-4x+4=9 \\\\ & \\Leftrightarrow {{(x-2)}^{2}}=9 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-2=3 \\\\ & x-2=-3 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=5 \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n})\\\\ & x=-1 \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n})\\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-1;5\\}$<span class='basic_pink'><br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$ v\u00e0 $5$<\/span> <\/span>"}]}],"id_ques":939},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[[1]]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv3/img\/8.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{{{x}^{2}}-x}+\\dfrac{1}{{{x}^{2}}-3x+2}+\\dfrac{1}{{{x}^{2}}-5x+6}+\\dfrac{1}{{{x}^{2}}-7x+12}=-1$ c\u00f3 t\u1eadp nghi\u1ec7m $S=\\varnothing$","select":["\u0110\u00fang","Sai"],"hint":"Ph\u00e2n t\u00edch m\u1ed7i m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed. \u00c1p d\u1ee5ng ph\u00e9p bi\u1ebfn \u0111\u1ed5i $\\dfrac {a}{x(x+a)}=\\dfrac {1}{x}-\\dfrac {1}{x+a}$","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $\\left\\{ \\begin{aligned} & {{x}^{2}}-x\\ne 0 \\\\ & {{x}^{2}}-3x+2\\ne 0 \\\\ & {{x}^{2}}-5x+6\\ne 0 \\\\ & {{x}^{2}}-7x+12\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x(x-1)\\ne 0 \\\\ & (x-1)(x-2)\\ne 0 \\\\ & (x-2)(x-3)\\ne 0 \\\\ & (x-3)(x-4)\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow\\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x\\ne 1 \\\\ & x\\ne 2 \\\\ & x\\ne 3 \\\\ & x\\ne 4 \\\\ \\end{aligned} \\right.$ <br\/>Ta c\u00f3:<br\/> $\\begin{aligned}& \\dfrac{1}{{{x}^{2}}-x}+\\dfrac{1}{{{x}^{2}}-3x+2}+\\dfrac{1}{{{x}^{2}}-5x+6}+\\dfrac{1}{{{x}^{2}}-7x+12}=-1\\\\ & \\Leftrightarrow \\dfrac{-1}{{{x}^{2}}-x}+\\dfrac{-1}{{{x}^{2}}-3x+2}+\\dfrac{-1}{{{x}^{2}}-5x+6}+\\dfrac{-1}{{{x}^{2}}-7x+12}=1 \\\\ & \\Leftrightarrow \\dfrac{-1}{x(x-1)}+\\dfrac{-1}{(x-1)(x-2)}+\\dfrac{-1}{(x-2)(x-3)}+\\dfrac{-1}{(x-3)(x-4)}=1 \\\\ & \\Leftrightarrow \\dfrac{1}{x}-\\dfrac{1}{x-1}+\\dfrac{1}{x-1}-\\dfrac{1}{x-2}+\\dfrac{1}{x-2}-\\dfrac{1}{x-3}+\\dfrac{1}{x-3}-\\dfrac{1}{x-4}=1 \\\\ & \\Leftrightarrow \\dfrac{1}{x}-\\dfrac{1}{x-4}=1 \\\\ & \\Leftrightarrow \\dfrac{x-4}{x(x-4)}-\\dfrac{x}{x(x-4)}=\\dfrac{x(x-4)}{x(x-4)} \\\\ & \\Rightarrow x-4-x=x(x-4) \\\\ & \\Leftrightarrow {{x}^{2}}-4x+4=0 \\\\ & \\Leftrightarrow {{(x-2)}^{2}}=0 \\\\ & \\Leftrightarrow x=2\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\varnothing$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang<\/span> <\/span>","column":2}]}],"id_ques":940}],"lesson":{"save":0,"level":3}}