{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/2.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{3x+2}{3x-2}=\\dfrac{9{{x}^{2}}}{9{{x}^{2}}-4}$ l\u00e0:","select":["A. $x\\ne \\dfrac {2}{3}$","B. $x\\ne 0$ v\u00e0 $x\\ne \\dfrac {2}{3}$","C. $x\\ne \\dfrac {2}{3}$ v\u00e0 $x\\ne \\dfrac {-2}{3}$","D. $x\\ne \\dfrac {2}{3}$ ho\u1eb7c $x\\ne \\dfrac {-2}{3}$."],"hint":"T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $x$ \u0111\u1ec3 t\u1ea5t c\u1ea3 c\u00e1c m\u1eabu th\u1ee9c \u0111\u1ec1u kh\u00e1c $0$","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{3x+2}{3x-2}=\\dfrac{9{{x}^{2}}}{9{{x}^{2}}-4}$ l\u00e0:<br\/>$\\begin{aligned} & \\left\\{ \\begin{aligned} & 3x-2\\ne 0 \\\\ & 9{{x}^{2}}-4\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & 3x-2\\ne 0 \\\\ & (3x-2)(3x+2)\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne \\dfrac{2}{3} \\\\ & x\\ne \\dfrac{-2}{3} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":911},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/12.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{5x}{{{x}^{2}}+2x+1}+\\dfrac{x(x-1)}{{{x}^{3}}+1}=0$ l\u00e0:","select":["A. $x\\ne 0$ v\u00e0 $x\\ne 1$","B. $x\\in \\mathbb R$","C. $x\\ne -1$","D. $x\\ne 1$ v\u00e0 $x\\ne -1.$"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{5x}{{{x}^{2}}+2x+1}+\\dfrac{x(x-1)}{{{x}^{3}}+1}=0$ l\u00e0:<br\/>$\\begin{aligned} & \\left\\{ \\begin{aligned} & {{x}^{2}}+2x+1\\ne 0 \\\\ & {{x}^{3}}+1\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & {{(x+1)}^{2}}\\ne 0 \\\\ &(x+1)({{x}^{2}}-x+1)\\ne 0 \\\\ \\end{aligned} \\right. \\\\ &\\Leftrightarrow \\left\\{ \\begin{aligned} & x+1\\ne 0 \\\\ & {{x}^{2}}-x+1 \\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne -1 \\\\ & {{\\left( x-\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{3}{4}\\ne 0\\,\\,\\,\\,\\,\\,\\forall x \\\\ \\end{aligned} \\right. \\\\ &\\Leftrightarrow x\\ne -1 \\\\ \\end{aligned}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":912},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/10.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{3}{{{x}^{2}}-6x+5}=\\dfrac{1}{{{x}^{2}}+2}$ l\u00e0 $\\left\\{ \\begin{aligned}& x\\ne 1 \\\\ & x\\ne 5 \\\\ \\end{aligned} \\right. $","select":["\u0110\u00fang","Sai"],"hint":"T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $x$ \u0111\u1ec3 t\u1ea5t c\u1ea3 c\u00e1c m\u1eabu th\u1ee9c \u0111\u1ec1u kh\u00e1c $0$","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{3}{{{x}^{2}}-6x+5}=\\dfrac{1}{{{x}^{2}}+2}$ l\u00e0:<br\/>$\\begin{aligned} & \\left\\{ \\begin{aligned} & {{x}^{2}}-6x+5\\ne 0 \\\\ & {{x}^{2}}+2\\ne 0 \\\\ \\end{aligned} \\right. \\\\& \\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}^{2}}-5x-x+5\\ne 0 \\\\ & {{x}^{2}}+2\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & (x-1)(x-5)\\ne 0 \\\\ & {{x}^{2}}\\ne -2\\,\\,\\,\\,\\,\\,\\forall x \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned}& x\\ne 1 \\\\ & x\\ne 5 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang<\/span><\/span>","column":2}]}],"id_ques":913},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/13.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-1}{{{x}^{2}}-x+1}-\\dfrac{x}{{{x}^{2}}+x+1}=\\dfrac{1}{2{{x}^{2}}+1}$ l\u00e0:","select":["A. $x\\ne 0$ v\u00e0 $x\\ne 1$","B. $x\\in \\mathbb R$","C. $x\\ne \\dfrac {1}{4}$","D. $x\\ne -1;\\,x\\ne \\dfrac {1}{4}$ v\u00e0 $x\\ne 1$"],"hint":"T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $x$ \u0111\u1ec3 t\u1ea5t c\u1ea3 c\u00e1c m\u1eabu th\u1ee9c \u0111\u1ec1u kh\u00e1c $0$","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-1}{{{x}^{2}}-x+1}-\\dfrac{x}{{{x}^{2}}+x+1}=\\dfrac{1}{2{{x}^{2}}+1}$ l\u00e0:<br\/>$\\begin{aligned}& \\left\\{ \\begin{aligned}& {{x}^{2}}-x+1\\ne 0 \\\\ & {{x}^{2}}+x+1\\ne 0 \\\\ & 2{{x}^{2}}+1\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned}& {{\\left( x-\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{3}{4}\\ne 0 \\\\ & {{\\left( x+\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{3}{4}\\ne 0 \\\\ & {{x}^{2}}\\ne \\dfrac{-1}{2} \\\\ \\end{aligned} \\right.\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\forall x\\in \\mathbb R \\\\ \\end{aligned}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":914},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/4.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{x-1}-\\dfrac{7}{x-2}=\\dfrac{1}{(x-1)(2-x)}$ c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{1\\}.$","select":["\u0110\u00fang","Sai"],"hint":"Quy \u0111\u1ed3ng, kh\u1eed m\u1eabu v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu.","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1$ v\u00e0 $x\\ne 2$<br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\dfrac{1}{x-1}-\\dfrac{7}{x-2}=\\dfrac{1}{(x-1)(2-x)} \\\\ & \\Leftrightarrow \\dfrac{x-2}{x-1}-\\dfrac{7(x-1)}{x-2}=\\dfrac{-1}{(x-1)(x-2)} \\\\ & \\Rightarrow x-2-7(x-1)=-1 \\\\ & \\Leftrightarrow x-2-7x+7+1=0 \\\\ & \\Leftrightarrow -6x+6=0 \\\\ & \\Leftrightarrow x=1\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ \\end{align}$<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 Sai.<\/span><\/span>","column":2}]}],"id_ques":915},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/9.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x+1}{x-3}-\\dfrac{1}{x-1}=\\dfrac{2}{{{x}^{2}}-4x+3}$ c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{0;1\\}.$","select":["\u0110\u00fang","Sai"],"hint":"Ph\u00e2n t\u00edch m\u1eabu th\u1ee9c v\u1ebf ph\u1ea3i th\u00e0nh nh\u00e2n t\u1eed, r\u1ed3i quy \u0111\u1ed3ng m\u1eabu th\u1ee9c v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1$ v\u00e0 $x\\ne 3$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{x+1}{x-3}-\\dfrac{1}{x-1}=\\dfrac{2}{{{x}^{2}}-4x+3} \\\\ & \\Leftrightarrow \\dfrac{x+1}{x-3}-\\dfrac{1}{x-1}=\\dfrac{2}{x^2-x-3x+3} \\\\& \\Leftrightarrow \\dfrac{x+1}{x-3}-\\dfrac{1}{x-1}=\\dfrac{2}{x(x-1)-3(x-1)} \\\\ & \\Leftrightarrow \\dfrac{\\left( x+1 \\right)\\left( x-1 \\right)}{(x-1)(x-3)}-\\dfrac{x-3}{(x-1)(x-3)}=\\dfrac{2}{(x-1)(x-3)} \\\\ & \\Rightarrow \\left( x+1 \\right)\\left( x-1 \\right)-(x-3)=2 \\\\ & \\Leftrightarrow {{x}^{2}}-1-x+3-2=0 \\\\ & \\Leftrightarrow {{x}^{2}}-x=0 \\\\ & \\Leftrightarrow x(x-1)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=1 \\,\\,\\,\\,\\,(\\text{lo\u1ea1i})\\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{0\\}$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai.<\/span><\/span>","column":2}]}],"id_ques":916},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/8.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $x+\\dfrac{x-1}{x-2}=3+\\dfrac{1}{x-2}$ c\u00f3 t\u1eadp nghi\u1ec7m $S=\\varnothing .$","select":["\u0110\u00fang","Sai"],"hint":"Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c, r\u1ed3i kh\u1eed m\u1eabu v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2$<br\/>Ta c\u00f3:<br\/>$\\begin{align} & x+\\dfrac{x-1}{x-2}=3+\\dfrac{1}{x-2} \\\\ & \\Leftrightarrow \\dfrac{x(x-2)+x-1}{x-2}=\\dfrac{3(x-2)+1}{x-2} \\\\ & \\Rightarrow {{x}^{2}}-2x+x-1=3x-6+1 \\\\ & \\Leftrightarrow {{x}^{2}}-4x+4=0 \\\\ & \\Leftrightarrow {{(x-2)}^{2}}=0 \\\\ & \\Leftrightarrow x=2\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ \\end{align}$<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><br\/><i>Ch\u00fa \u00fd:<\/i> Sau khi t\u00ecm \u0111\u01b0\u1ee3c c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a $x$ ph\u1ea3i x\u00e9t xem c\u00e1c gi\u00e1 tr\u1ecb \u0111\u00f3 c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh ph\u01b0\u01a1ng tr\u00ecnh kh\u00f4ng. <\/span>","column":2}]}],"id_ques":917},{"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"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/11.jpg' \/><\/center><br\/>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-4}{x(x+2)}+\\dfrac{2}{{{x}^{2}}-4}=\\dfrac{1}{x(x-2)}.$<br\/> T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$ ","hint":"S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$, ph\u00e2n t\u00edch m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i quy \u0111\u1ed3ng v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 0$ v\u00e0 $x\\ne \\pm 2$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{x-4}{x(x+2)}+\\dfrac{2}{{{x}^{2}}-4}=\\dfrac{1}{x(x-2)} \\\\ & \\Leftrightarrow \\dfrac{(x-4)(x-2)}{x(x-2)(x+2)}+\\dfrac{2x}{x(x-2)(x+2)}=\\dfrac{x+2}{x(x-2)(x+2)} \\\\ & \\Rightarrow (x-4)(x-2)+2x=x+2 \\\\ & \\Leftrightarrow {{x}^{2}}-6x+8+2x-x-2=0 \\\\ & \\Leftrightarrow {{x}^{2}}-5x+6=0 \\\\ &\\Leftrightarrow x^2-2x-3x+6=0\\\\ &\\Leftrightarrow x(x-2)-3(x-2)=0\\\\ & \\Leftrightarrow (x-2)(x-3)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ & x=3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{3\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $3$<\/span><br\/><i><span class='basic_green'>Ch\u00fa \u00fd:<\/span><\/i> Khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, tr\u01b0\u1edbc khi quy \u0111\u1ed3ng ta n\u00ean th\u1ef1c hi\u1ec7n ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed \u0111\u1ec3 d\u1ec5 d\u00e0ng x\u00e1c \u0111\u1ecbnh m\u1eabu th\u1ee9c chung h\u01a1n. \u0110\u1ed3ng th\u1eddi, \u0111\u1ec3 m\u1eabu th\u1ee9c chung l\u00e0 \u0111\u01a1n gi\u1ea3n nh\u1ea5t gi\u00fap c\u00e1c ph\u00e9p bi\u1ebfn \u0111\u1ed5i sau d\u1ec5 d\u00e0ng h\u01a1n.<\/span>"}]}],"id_ques":918},{"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/lv2/img\/12.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-1}{x+3}-\\dfrac{x}{x-3}=\\dfrac{7x-3}{9-{{x}^{2}}}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{0\\}$","B. $S=\\mathbb R$","C. $S=\\varnothing$","D. $S=\\{1\\}$"],"hint":"S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$, ph\u00e2n t\u00edch m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i quy \u0111\u1ed3ng v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne \\pm 3$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{x-1}{x+3}-\\dfrac{x}{x-3}=\\dfrac{7x-3}{9-{{x}^{2}}} \\\\ & \\Leftrightarrow \\dfrac{x-1}{x+3}-\\dfrac{x}{x-3}=\\dfrac{3-7x}{{{x}^{2}}-9} \\\\ & \\Leftrightarrow \\dfrac{(x-1)(x-3)}{(x-3)(x+3)}-\\dfrac{x(x+3)}{(x-3)(x+3)}=\\dfrac{3-7x}{(x-3)(x+3)} \\\\ & \\Rightarrow (x-1)(x-3)-x(x+3)=3-7x \\\\ & \\Leftrightarrow {{x}^{2}}-4x+3-{{x}^{2}}-3x-3+7x=0 \\\\ & \\Leftrightarrow 0x=0\\,\\,\\,\\,\\,\\,\\,(\\text{lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi }x\\in\\mathbb R) \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\mathbb R$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":919},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/11.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x}{x-5}-\\dfrac{3}{x-2}=1.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{\\dfrac{-5}{2}\\right\\}$","B. $S=\\left\\{\\dfrac{5}{2}\\right\\}$","C. $S=\\left\\{\\dfrac{2}{5}\\right\\}$","D. $S=\\left\\{\\dfrac{-2}{5}\\right\\}$"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2$ v\u00e0 $x\\ne 5$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{x}{x-5}-\\dfrac{3}{x-2}=1 \\\\ & \\Leftrightarrow \\dfrac{x(x-2)}{(x-5)(x-2)}-\\dfrac{3(x-5)}{(x-5)(x-2)}=\\dfrac{(x-5)(x-2)}{(x-5)(x-2)} \\\\ &\\Rightarrow x(x-2)-3(x-5)=(x-5)(x-2) \\\\ & \\Leftrightarrow {{x}^{2}}-2x-3x+15={{x}^{2}}-7x+10 \\\\ & \\Leftrightarrow 2x=-5 \\\\ & \\Leftrightarrow x=\\dfrac{-5}{2} \\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n})\\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\dfrac{-5}{2}\\right\\}.$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":920},{"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/lv2/img\/13.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{x-1}+\\dfrac{1}{x-2}=\\dfrac{3}{2x-6}-\\dfrac{1}{2x-4}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{\\dfrac{2}{3};5\\right\\}$","B. $S=\\left\\{\\dfrac{3}{2};5\\right\\}$","C. $S=\\left\\{\\dfrac{3}{2}\\right\\}$","D. $S=\\left\\{\\dfrac{-3}{2};5\\right\\}$"],"hint":"Ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i quy \u0111\u1ed3ng v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1;x\\ne 2$ v\u00e0 $x\\ne 3$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{1}{x-1}+\\dfrac{1}{x-2}=\\dfrac{3}{2x-6}-\\dfrac{1}{2x-4} \\\\ & \\Leftrightarrow \\dfrac{1}{x-1}+\\dfrac{1}{x-2}=\\dfrac{3}{2(x-3)}-\\dfrac{1}{2(x-2)} \\\\ & \\Leftrightarrow \\dfrac{2(x-2)(x-3)}{(x-1)(x-2)(x-3)}+\\dfrac{2(x-1)(x-3)}{(x-1)(x-2)(x-3)}=\\dfrac{3(x-2)(x-1)}{(x-1)(x-2)(x-3)}-\\dfrac{(x-1)(x-3)}{(x-1)(x-2)(x-3)} \\\\ & \\Rightarrow 2(x-2)(x-3)+2(x-1)(x-3)=3(x-2)(x-1)-(x-1)(x-3)\\\\&\\Leftrightarrow 2{{x}^{2}}-10x+12+2{{x}^{2}}-8x+6=3{{x}^{2}}-9x+6-{{x}^{2}}+4x-3 \\\\ &\\Leftrightarrow 4{{x}^{2}}-18x+18=2{{x}^{2}}-5x+3 \\\\ & \\Leftrightarrow 2{{x}^{2}}-13x+15=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}-10x-3x+15=0 \\\\ & \\Leftrightarrow 2x(x-5)-3(x-5)=0 \\\\ & \\Leftrightarrow (2x-3)(x-5)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{3}{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\dfrac{3}{2};5\\right\\}.$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":921},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/9.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{x+2}-\\dfrac{x-5}{x-1}+1=\\dfrac{3}{{{x}^{2}}+x-2}$.<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{\\dfrac{-4}{5}\\right\\}$","B. $S=\\left\\{\\dfrac{4}{5}\\right\\}$","C. $S=\\left\\{\\dfrac{-5}{4}\\right\\}$","D. $S=\\varnothing$"],"hint":"Ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i quy \u0111\u1ed3ng v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1;x\\ne -2$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{1}{x+2}-\\dfrac{x-5}{x-1}+1=\\dfrac{3}{{{x}^{2}}+x-2} \\\\ & \\Leftrightarrow \\dfrac{1}{x+2}-\\dfrac{x-5}{x-1}+1=\\dfrac{3}{(x-1)(x+2)} \\\\ & \\Leftrightarrow \\dfrac{x-1}{(x-1)(x+2)}-\\dfrac{(x-5)(x+2)}{(x-1)(x+2)}+\\dfrac{(x-1)(x+2)}{(x-1)(x+2)}=\\dfrac{3}{(x-1)(x+2)} \\\\ & \\Rightarrow x-1-(x-5)(x+2)+(x-1)(x+2)=3 \\\\ & \\Leftrightarrow x-1-({{x}^{2}}-3x-10)+({{x}^{2}}+x-2)=3 \\\\ & \\Leftrightarrow x-1-{{x}^{2}}+3x+10+{{x}^{2}}+x-2-3=0 \\\\ & \\Leftrightarrow 5x+4=0 \\\\ & \\Leftrightarrow x=\\dfrac{-4}{5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\dfrac{-4}{5}\\right\\}.$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":922},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\left[\\begin{array}{l}{a=0} \\\\ {a = \\dfrac{-9}{2}}\\end{array}\\right.$","B. $\\left[\\begin{array}{l}{a= 2} \\\\ {a = \\dfrac{-1}{2}}\\end{array}\\right.$","C. $\\left[\\begin{array}{l}{a=1} \\\\ {a = \\dfrac{-1}{2}}\\end{array}\\right.$"],"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-a}{x+a}-\\dfrac{x+a}{x-a}+\\dfrac{2{{a}^{2}}+a}{{{x}^{2}}-{{a}^{2}}}=0$ ($x$ l\u00e0 \u1ea9n, $a$ l\u00e0 tham s\u1ed1). T\u00ecm $a$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=-2$<\/span> ","hint":"Thay gi\u00e1 tr\u1ecb $x=-2$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $a$.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\pm a$.<br\/>Thay $x=-2$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u. Ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{(-2)-a}{(-2)+a}-\\dfrac{(-2)+a}{(-2)-a}+\\dfrac{2{{a}^{2}}+a}{{{(-2)}^{2}}-{{a}^{2}}}=0 \\\\ & \\Leftrightarrow \\dfrac{-\\left( a+2 \\right)}{a-2}-\\dfrac{a-2}{-\\left( a+2 \\right)}+\\dfrac{2{{a}^{2}}+a}{-\\left( {{a}^{2}}-4 \\right)}=0 \\\\ & \\Leftrightarrow -\\dfrac{a+2}{a-2}+\\dfrac{a-2}{a+2}-\\dfrac{2{{a}^{2}}+a}{(a-2)(a+2)}=0 \\\\ & \\Leftrightarrow \\dfrac{{{-(a+2)}^{2}}}{(a-2)(a+2)}+\\dfrac{{{(a-2)}^{2}}}{(a-2)(a+2)}-\\dfrac{2{{a}^{2}}+a}{(a-2)(a+2)}=0 \\\\ & \\Rightarrow -({{a}^{2}}+4a+4)+({{a}^{2}}-4a+4)-2{{a}^{2}}-a=0 \\\\ & \\Leftrightarrow -2a^2 -9a=0 \\\\ &\\Leftrightarrow 2a^2+9a=0\\\\ & \\Leftrightarrow a(2a+9)=0\\\\ & \\Leftrightarrow \\left[\\begin{aligned} & a=0\\\\ & 2a=-9\\\\ \\end{aligned}\\right.\\\\ & \\Leftrightarrow \\left[\\begin{aligned} & a=0\\\\ & a=\\dfrac{-9}{2}\\\\ \\end{aligned}\\right. \\\\ \\end{aligned}$<br\/>V\u1eady v\u1edbi $a=0$ ho\u1eb7c $a=\\dfrac{-9}{2}$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=-2$<\/span>"}]}],"id_ques":923},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/4.jpg' \/><\/center><br\/><span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{a}{x-a}+\\dfrac{2{{x}^{2}}-5}{{{x}^{3}}-a}=\\dfrac{4}{{{x}^{2}}+ax+1}$ ( $x$ l\u00e0 \u1ea9n, $a$ l\u00e0 tham s\u1ed1). Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi $a=1$. <br\/><b> \u0110\u00e1p s\u1ed1: <\/b>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span> ","hint":"Thay gi\u00e1 tr\u1ecb $a=1$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne a$.<br\/>Thay $a=1$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh. Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{1}{x-1}+\\dfrac{2{{x}^{2}}-5}{{{x}^{3}}-1}=\\dfrac{4}{{{x}^{2}}+x+1} \\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n:}x\\ne 1)\\\\ & \\Leftrightarrow \\dfrac{1}{x-1}+\\dfrac{2{{x}^{2}}-5}{(x-1)({{x}^{2}}+x+1)}=\\dfrac{4}{{{x}^{2}}+x+1} \\\\ & \\Leftrightarrow \\dfrac{{{x}^{2}}+x+1}{(x-1)({{x}^{2}}+x+1)}+\\dfrac{2{{x}^{2}}-5}{(x-1)({{x}^{2}}+x+1)}=\\dfrac{4(x-1)}{(x-1)({{x}^{2}}+x+1)} \\\\ & \\Rightarrow {{x}^{2}}+x+1+2{{x}^{2}}-5=4(x-1) \\\\ & \\Leftrightarrow 3{{x}^{2}}+x-4-4x+4=0 \\\\ & \\Leftrightarrow 3{{x}^{2}}-3x=0 \\\\ & \\Leftrightarrow 3x(x-1)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned}& x=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i v\u00ec b\u1eb1ng} a) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady v\u1edbi $a=1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{0\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$<\/span><\/span>"}]}],"id_ques":924},{"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/lv2/img\/13.jpg' \/><\/center>Cho hai bi\u1ec3u th\u1ee9c $A=\\dfrac {12}{x^3+8}$ v\u00e0 $B=\\dfrac {1}{x+2}+1$. T\u00ecm $x$ \u0111\u1ec3 $A=B.$","select":["A. $x=0$","B. $x\\in\\{0;1\\}$","C. $x\\in\\{0;1;2\\}$","D. $x\\in \\varnothing$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=B$","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & A=B \\\\ & \\Leftrightarrow \\dfrac{12}{{{x}^{3}}+8}=\\dfrac{1}{x+2}+1 \\,\\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n: } x\\ne -2)\\\\ & \\Leftrightarrow \\dfrac{12}{(x+2)({{x}^{2}}-2x+4)}=\\dfrac{{{x}^{2}}-2x+4}{(x+2)({{x}^{2}}-2x+4)}+\\dfrac{(x+2)({{x}^{2}}-2x+4)}{(x+2)({{x}^{2}}-2x+4)} \\\\ & \\Rightarrow 12={{x}^{2}}-2x+4+(x+2)({{x}^{2}}-2x+4) \\\\ & \\Leftrightarrow {{x}^{2}}-2x+4+{{x}^{3}}+8-12=0 \\\\ & \\Leftrightarrow {{x}^{3}}+{{x}^{2}}-2x=0 \\\\ & \\Leftrightarrow x({{x}^{2}}+x-2)=0 \\\\ & \\Leftrightarrow x(x-1)(x+2)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x-1=0 \\\\ & x+2=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n)} \\\\ & x=-2\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady $x\\in\\{0;1\\}$ th\u00ec $A=B$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":925},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/11.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{2}{{{x}^{2}}+2x+1}-\\dfrac{5}{{{x}^{2}}-2x+1}-\\dfrac{3}{1-{{x}^{2}}}$. T\u00ecm $x$ \u0111\u1ec3 $A=0.$","select":["A. $x=\\dfrac{7}{3}$","B. $x=\\dfrac{3}{7}$","C. $x=\\dfrac{-3}{7}$","D. $x=\\dfrac{-7}{3}$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=0$. \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed. ","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} &A=0\\\\&\\Leftrightarrow \\dfrac{2}{{{x}^{2}}+2x+1}-\\dfrac{5}{{{x}^{2}}-2x+1}-\\dfrac{3}{1-{{x}^{2}}}=0 \\\\ & \\Leftrightarrow \\dfrac{2}{{{(x+1)}^{2}}}-\\dfrac{5}{{{(x-1)}^{2}}}+\\dfrac{3}{(x-1)(x+1)}=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n: } x\\ne \\pm 1) \\\\ & \\Leftrightarrow \\dfrac{2{{(x-1)}^{2}}}{{{(x+1)}^{2}}{{(x-1)}^{2}}}-\\dfrac{5{{(x+1)}^{2}}}{{{(x+1)}^{2}}{{(x-1)}^{2}}}+\\dfrac{3(x-1)(x+1)}{{{(x+1)}^{2}}{{(x-1)}^{2}}}=0 \\\\ & \\Rightarrow 2{{(x-1)}^{2}}-5{{(x+1)}^{2}}+3(x-1)(x+1)=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}-4x+2-5{{x}^{2}}-10x-5+3{{x}^{2}}-3=0 \\\\ & \\Leftrightarrow -14x-6=0 \\\\ & \\Leftrightarrow x=\\dfrac{-3}{7}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned}$<br\/>V\u1eady $x=\\dfrac{-3}{7}$ th\u00ec $A=0$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":926},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/10.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac {1}{x-1}+\\dfrac {2}{x-2}=\\dfrac {3}{x-3}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{\\dfrac{3}{2}\\right\\}$","B. $S=\\left\\{\\dfrac{2}{3}\\right\\}$","C. $S=\\left\\{\\dfrac{-3}{2}\\right\\}$","D. $S=\\left\\{\\dfrac{-2}{3}\\right\\}$"],"hint":"Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c, kh\u1eed m\u1eabu \u0111\u01b0a v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch. ","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1; x\\ne 2$ v\u00e0 $x\\ne 3$<br\/>Ta c\u00f3:<br\/>$\\begin{align} & \\dfrac{1}{x-1}+\\dfrac{2}{x-2}=\\dfrac{3}{x-3} \\\\ & \\Leftrightarrow \\dfrac{(x-2)(x-3)}{(x-1)(x-2)(x-3)}+\\dfrac{2(x-1)(x-3)}{(x-1)(x-2)(x-3)}=\\dfrac{3(x-1)(x-2)}{(x-1)(x-2)(x-3)} \\\\ & \\Rightarrow (x-2)(x-3)+2(x-1)(x-3)=3(x-1)(x-2) \\\\ & \\Leftrightarrow ({{x}^{2}}-5x+6)+2({{x}^{2}}-4x+3)-3({{x}^{2}}-3x+2)=0 \\\\ & \\Leftrightarrow -4x+6=0 \\\\ & \\Leftrightarrow x=\\dfrac{3}{2} \\,\\,\\,(\\text{th\u1ecfa m\u00e3n})\\\\ \\end{align}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\dfrac{3}{2}\\right\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":927},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/10.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{5}{{{x}^{2}}+2x}+{{x}^{2}}+2x+6=0.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{-1;-5\\}$","B. $S=\\{1;5\\}$","C. $S=\\varnothing$","D. $S=\\{-1\\}$"],"hint":"\u0110\u1eb7t $x^2+2x=t$. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$ r\u1ed3i t\u00ecm $x$ ","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> \u0110\u1eb7t $x^2+2x=t$. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay gi\u00e1 tr\u1ecb c\u1ee7a t v\u00e0o $x^2+2x=t$, r\u1ed3i t\u00ecm $x$.<br\/><b>B\u01b0\u1edbc 4:<\/b> Lo\u1ea1i nghi\u1ec7m r\u1ed3i k\u1ebft lu\u1eadn<\/span> <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x^2+2x \\ne 0 \\Leftrightarrow x(x+2)\\ne 0 \\Leftrightarrow \\left\\{\\begin{aligned}& x\\ne 0\\\\ & x\\ne -2\\\\ \\end{aligned}\\right.$<br\/>\u0110\u1eb7t $x^2+2x=t$, ($t\\ne 0)$. <br\/>Ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned} & \\dfrac{5}{t}+t+6=0 \\\\ & \\Leftrightarrow \\dfrac{5}{t}+\\dfrac{{{t}^{2}}}{t}+\\dfrac{6t}{t}=0 \\\\ & \\Rightarrow {{t}^{2}}+6t+5=0 \\\\ & \\Leftrightarrow (t+1)(t+5)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t+1=0\\\\ & t+5=0 \\\\ \\end{aligned} \\right. \\\\& \\Leftrightarrow \\left[ \\begin{aligned} & t=-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & t=-5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1edbi $t=-1$. Ta c\u00f3:<br\/>$\\begin{align} & {{x}^{2}}+2x=-1 \\\\ & \\Leftrightarrow {{x}^{2}}+2x+1=0 \\\\ & \\Leftrightarrow {{(x+1)}^{2}}=0 \\\\ & \\Leftrightarrow x=-1\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{align}$<br\/>V\u1edbi $t=-5$, ta c\u00f3:<br\/>$\\begin{align}& {{x}^{2}}+2x=-5 \\\\ & \\Leftrightarrow {{x}^{2}}+2x+5=0 \\\\ & \\Leftrightarrow {{(x+1)}^{2}}=-4\\,\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m}) \\\\ \\end{align}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-1\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":928},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/12.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x^2+x+\\dfrac {1}{x^2}+\\dfrac {1}{x}=0.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{1;-2\\}$","B. $S=\\{-1;2\\}$","C. $S=\\varnothing$","D. $S=\\{-1\\}$"],"hint":"\u0110\u1eb7t $x+\\dfrac {1}{x}=t$. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$ r\u1ed3i t\u00ecm $x$ ","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> \u0110\u1eb7t $x+\\dfrac {1}{x}=t\\Rightarrow x^2+\\dfrac {1}{x^2}=t^2-2$. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay gi\u00e1 tr\u1ecb c\u1ee7a $t$ v\u00e0o $x+\\dfrac {1}{x}=t$, r\u1ed3i t\u00ecm $x$.<br\/><b>B\u01b0\u1edbc 4:<\/b> Lo\u1ea1i nghi\u1ec7m r\u1ed3i k\u1ebft lu\u1eadn<\/span> <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 0$<br\/>\u0110\u1eb7t $x+\\dfrac {1}{x}=t$ <br\/>Ta c\u00f3: ${{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{x}^{2}}+2.x.\\dfrac{1}{x}+\\dfrac{1}{{{x}^{2}}}-2={{\\left( x+\\dfrac{1}{x} \\right)}^{2}}-2={{t}^{2}}-2$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned} & {{t}^{2}}+t-2=0 \\\\ & \\Leftrightarrow t^2-t+2t-2=0\\\\ &\\Leftrightarrow t(t-1)+2(t-1)=0 \\\\ & \\Leftrightarrow (t-1)(t+2)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t-1=0 \\\\ & t+2=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=1 \\\\ & t=-2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1edbi $t=1$. Ta c\u00f3:<br\/>$\\begin{aligned} & x+\\dfrac{1}{x}=1 \\\\ & \\Leftrightarrow {{x}^{2}}-x+1=0 \\\\ & \\Leftrightarrow {{\\left( x-\\dfrac{1}{2}\\right)}^{2}}+\\dfrac{3}{4}=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m}) \\\\ \\end{aligned}$<br\/>V\u1edbi $t=-2$, ta c\u00f3:<br\/>$\\begin{align} & x+\\dfrac{1}{x}=-2 \\\\ & \\Leftrightarrow {{x}^{2}}+2x+1=0 \\\\ & \\Leftrightarrow {{(x+1)}^{2}}=0 \\\\ & \\Leftrightarrow x=-1\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{align}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-1\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":929},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv2/img\/12.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x\\left( \\dfrac{1}{x}-\\dfrac{2}{{{x}^{2}}} \\right)+\\dfrac{1}{x}\\left( 1-\\dfrac{6}{x} \\right)=0.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{-2;3\\}$","B. $S=\\{2;-3\\}$","C. $S=\\varnothing$","D. $S=\\{-2\\}$"],"hint":"Th\u1ef1c hi\u1ec7n ph\u00e9p nh\u00e2n v\u00e0 quy \u0111\u1ed3ng \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch. ","explain":" <span class='basic_left'><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 0$<br\/>Ta c\u00f3: <br\/>$\\begin{aligned} & x\\left( \\dfrac{1}{x}-\\dfrac{2}{{{x}^{2}}} \\right)+\\dfrac{1}{x}\\left( 1-\\dfrac{6}{x} \\right)=0 \\\\ & \\Leftrightarrow x.\\dfrac{1}{x}-\\dfrac{2x}{{{x}^{2}}}+\\dfrac{1}{x}-\\dfrac{6}{{{x}^{2}}}=0 \\\\ & \\Leftrightarrow 1-\\dfrac{2}{x}+\\dfrac{1}{x}-\\dfrac{6}{{{x}^{2}}}=0 \\\\ & \\Leftrightarrow \\dfrac{{{x}^{2}}}{{{x}^{2}}}-\\dfrac{x}{{{x}^{2}}}-\\dfrac{6}{{{x}^{2}}}=0 \\\\ & \\Leftrightarrow {{x}^{2}}-x-6=0 \\\\ &\\Leftrightarrow x^2+2x-3x-6=0\\\\ & \\Leftrightarrow x(x+2)-3(x+2)=0\\\\ & \\Leftrightarrow (x+2)(x-3)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned}& x=-2\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=3\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-2;3\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":930}],"lesson":{"save":0,"level":2}}