{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank_random","correct":[[["5"],["-5"],["0"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/12.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{x+2}{{{x}^{2}}-5x}+\\dfrac{x-2}{{{x}^{2}}+5x} \\right)$$:\\dfrac{{{x}^{2}}+10}{{{x}^{2}}-25} $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $C$ l\u00e0: $\\left\\{ \\begin{aligned} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","hint":" \u0110\u1ec3 ph\u00e2n th\u1ee9c c\u00f3 ngh\u0129a th\u00ec m\u1eabu th\u1ee9c ph\u1ea3i kh\u00e1c 0 ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $C$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x+5\\ne 0 \\\\ & x-5\\ne 0 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -5 \\\\ & x\\ne 5 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $5; -5$ v\u00e0 $0.$ <\/span><\/span> "}]}],"id_ques":151},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["2"],["x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/12.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{x+2}{{{x}^{2}}-5x}+\\dfrac{x-2}{{{x}^{2}}+5x} \\right)$$:\\dfrac{{{x}^{2}}+10}{{{x}^{2}}-25} $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $C$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $C = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $x\\ne \\{-5;0;5\\}$ <br\/> Ta c\u00f3: <br\/> $C=\\left( \\dfrac{x+2}{{{x}^{2}}-5x}+\\dfrac{x-2}{{{x}^{2}}+5x} \\right)$$:\\dfrac{{{x}^{2}}+10}{{{x}^{2}}-25} $<br\/>$ =\\left[ \\dfrac{x+2}{x\\left( x-5 \\right)}+\\dfrac{x-2}{x\\left( x+5 \\right)} \\right]$$:\\dfrac{{{x}^{2}}+10}{{{x}^{2}}-25} $<br\/>$ = \\dfrac{\\left( x+2 \\right)\\left( x+5 \\right)+\\left( x-2 \\right)\\left( x-5 \\right)}{x\\left( {{x}^{2}}-25 \\right)} $$\\cdot \\dfrac{{{x}^{2}}-25}{{{x}^{2}}+10} $<br\/>$ =\\dfrac{{{x}^{2}}+7x+10+{{x}^{2}}-7x+10}{x\\left( {{x}^{2}}-25 \\right)}\\cdot \\dfrac{{{x}^{2}}-25}{{{x}^{2}}+10} $<br\/>$ =\\dfrac{2{{x}^{2}}+20}{x}\\cdot \\dfrac{1}{{{x}^{2}}+10}$<br\/>$ =\\dfrac{2\\left( {{x}^{2}}+10 \\right)}{x}\\cdot \\dfrac{1}{{{x}^{2}}+10} $<br\/>$ =\\dfrac{2}{x}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $\\dfrac{2}{x}$. <\/span><\/span> "}]}],"id_ques":152},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["1"],["2017"]]],"list":[{"point":5,"width":60,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/12.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{x+2}{{{x}^{2}}-5x}+\\dfrac{x-2}{{{x}^{2}}+5x} \\right)$$:\\dfrac{{{x}^{2}}+10}{{{x}^{2}}-25} $<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $x=4034$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $C$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":" Thay $x=4034$ v\u00e0o bi\u1ec3u th\u1ee9c $C$ \u0111\u00e3 r\u00fat g\u1ecdn v\u00e0 t\u00ednh. ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $C=\\dfrac{2}{x}$ <br\/> Thay $x=4034$ v\u00e0o bi\u1ec3u th\u1ee9c $C$ \u0111\u00e3 r\u00fat g\u1ecdn tr\u00ean, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{2}{x}=\\dfrac{2}{4034}$$=\\dfrac{1}{2017}$<\/span> "}]}],"id_ques":153},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/12.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{x+2}{{{x}^{2}}-5x}+\\dfrac{x-2}{{{x}^{2}}+5x} \\right)$$:\\dfrac{{{x}^{2}}+10}{{{x}^{2}}-25} $ <br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> Kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $x$ \u0111\u1ec3 $C=2$ <\/span> ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $C=\\dfrac{2}{x}$ <br\/> $C=2\\Leftrightarrow$ $\\dfrac{2}{x}=2$<br\/> $\\Leftrightarrow x=1$ (th\u1ecfa m\u00e3n) <br\/> <b>K\u1ebft lu\u1eadn:<\/b> \u0110\u1ec3 $C=2$ th\u00ec $x=1$ <span><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":154},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank_random","correct":[[["2"],["-2"],["0"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $P=\\left( \\dfrac{1}{{{x}^{2}}+4x+4}-\\dfrac{1}{{{x}^{2}}-4x+4} \\right)$$:\\left( \\dfrac{1}{x+2}+\\dfrac{1}{x-2} \\right)$ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $P$ l\u00e0: $\\left\\{ \\begin{aligned} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\\\ & x \\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $P$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned}& {{x}^{2}}+4x+4\\ne 0 \\\\ & {{x}^{2}}-4x+4\\ne 0 \\\\ & x-2\\ne 0 \\\\ & x+2\\ne 0 \\\\ & \\frac{1}{x+2}+\\frac{1}{x-2}\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & {{\\left( x+2 \\right)}^{2}}\\ne 0 \\\\ & {{\\left( x-2 \\right)}^{2}}\\ne 0 \\\\ & x\\ne 2 \\\\ & x\\ne -2 \\\\ & \\frac{2x}{{{x}^{2}}-4}\\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 2 \\\\ & x\\ne -2 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$; $-2$ v\u00e0 $0$. <\/span><\/span> "}]}],"id_ques":155},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["-4"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $P=\\left( \\dfrac{1}{{{x}^{2}}+4x+4}-\\dfrac{1}{{{x}^{2}}-4x+4} \\right)$$:\\left( \\dfrac{1}{x+2}+\\dfrac{1}{x-2} \\right)$ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $P$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $P = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{x^2-4}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $x\\ne \\{-2;2; 0\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & P=\\left( \\dfrac{1}{{{x}^{2}}+4x+4}-\\dfrac{1}{{{x}^{2}}-4x+4} \\right):\\left( \\dfrac{1}{x+2}+\\dfrac{1}{x-2} \\right) \\\\ & =\\left[ \\dfrac{1}{{{\\left( x+2 \\right)}^{2}}}-\\dfrac{1}{{{\\left( x-2 \\right)}^{2}}} \\right]:\\left( \\dfrac{1}{x+2}+\\dfrac{1}{x-2} \\right) \\\\ & =\\left[ \\left( \\dfrac{1}{x+2}+\\dfrac{1}{x-2} \\right)\\left( \\dfrac{1}{x+2}-\\dfrac{1}{x-2} \\right) \\right]:\\left( \\dfrac{1}{x+2}+\\dfrac{1}{x-2} \\right) \\\\ & =\\dfrac{1}{x+2}-\\dfrac{1}{x-2} \\\\ & =\\dfrac{x-2}{\\left( x+2 \\right)\\left( x-2 \\right)}-\\dfrac{x+2}{\\left( x+2 \\right)\\left( x-2 \\right)} \\\\ & =\\dfrac{x-2-x-2}{\\left( x+2 \\right)\\left( x-2 \\right)} \\\\ & =\\dfrac{-4}{{{x}^{2}}-4} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-4.$ <\/span><\/span> "}]}],"id_ques":156},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["-1"],["8"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/10.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $P=\\left( \\dfrac{1}{{{x}^{2}}+4x+4}-\\dfrac{1}{{{x}^{2}}-4x+4} \\right)$$:\\left( \\dfrac{1}{x+2}+\\dfrac{1}{x-2} \\right)$ <br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $x=6$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $P$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $P=\\dfrac{-4}{x^2-4}$ <br\/> Thay $x=6$ v\u00e0o bi\u1ec3u th\u1ee9c $P$ \u0111\u00e3 r\u00fat g\u1ecdn tr\u00ean, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{-4}{x^2-4}=\\dfrac{-4}{6^2-4}=\\dfrac{-1}{8}$<\/span> "}]}],"id_ques":157},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $P=\\left( \\dfrac{1}{{{x}^{2}}+4x+4}-\\dfrac{1}{{{x}^{2}}-4x+4} \\right)$$:\\left( \\dfrac{1}{x+2}+\\dfrac{1}{x-2} \\right)$ <br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> Ta lu\u00f4n c\u00f3 $P < 0$ v\u1edbi m\u1ecdi $x\\ne \\{-2;2; 0\\}$ <\/span> ","select":["\u0110\u00fang","Sai"],"hint":" Ta nh\u1eadn \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a t\u1eed th\u1ee9c v\u00e0 m\u1eabu th\u1ee9c c\u1ee7a bi\u1ec3u th\u1ee9c $P$ \u0111\u00e3 r\u00fat g\u1ecdn. ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $P=\\dfrac{-4}{x^2-4}$ <br\/> Ta nh\u1eadn th\u1ea5y t\u1eed th\u1ee9c c\u1ee7a $P$ l\u00e0 $-4 < 0$<br\/> M\u1eabu th\u1ee9c c\u1ee7a $P$ l\u00e0 $x^2-4$ ch\u01b0a x\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb v\u1edbi m\u1ecdi $x\\ne \\{-2;2; 0\\}$.<br\/> Do \u0111\u00f3 k\u1ebft lu\u1eadn $P < 0$ v\u1edbi m\u1ecdi $x\\ne \\{-2;2; 0\\}$ l\u00e0 <b> sai<\/b>. <span><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":158},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank_random","correct":[[["1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\dfrac{x+2}{{{x}^{2}}+x+1}-\\dfrac{2}{x-1}$$-\\dfrac{2{{x}^{2}}+4}{1-{{x}^{3}}} $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $Q$ l\u00e0: $x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $Q$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x-1\\ne 0 \\\\ & x^2+x+1\\ne 0 \\\\ & 1 - x^3 \\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 1 \\\\ & x^2+x+1\\ne 0 \\,\\,\\forall x \\\\ \\end{aligned} \\right. \\Rightarrow x \\ne 1$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1.$ <\/span><\/span> "}]}],"id_ques":159},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\dfrac{x+2}{{{x}^{2}}+x+1}-\\dfrac{2}{x-1}$$-\\dfrac{2{{x}^{2}}+4}{1-{{x}^{3}}} $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $Q$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $Q = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{x^2+x+1}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $x\\ne 1$ <br\/> Ta c\u00f3: <br\/> $Q=\\dfrac{x+2}{{{x}^{2}}+x+1}-\\dfrac{2}{x-1}$$-\\dfrac{2{{x}^{2}}+4}{1-{{x}^{3}}} $<br\/>$ =\\dfrac{\\left( x+2 \\right)\\left( x-1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)}$$-\\dfrac{2\\left( {{x}^{2}}+x+1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)}$$+\\dfrac{2{{x}^{2}}+4}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $<br\/>$ =\\dfrac{\\left( x+2 \\right)\\left( x-1 \\right)-2\\left( {{x}^{2}}+x+1 \\right)+2{{x}^{2}}+4}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $<br\/>$ =\\dfrac{{{x}^{2}}+x-2-2{{x}^{2}}-2x-2+2{{x}^{2}}+4}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $<br\/>$ =\\dfrac{{{x}^{2}}-x}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $<br\/>$ =\\dfrac{x\\left( x-1 \\right)}{\\left( x-1 \\right)\\left( {{x}^{2}}+x+1 \\right)} $<br\/>$ =\\dfrac{x}{{{x}^{2}}+x+1}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $x$. <\/span><\/span> "}]}],"id_ques":160},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\dfrac{x+2}{{{x}^{2}}+x+1}-\\dfrac{2}{x-1}$$-\\dfrac{2{{x}^{2}}+4}{1-{{x}^{3}}} $<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x$ \u0111\u1ec3 $Q=0$<br\/> \u0110\u00e1p \u00e1n: $x = $_input_ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $Q=\\dfrac{x}{x^2+x+1}$ <br\/> $Q=0\\Leftrightarrow$$\\dfrac{x}{x^2+x+1}=0\\Leftrightarrow x=0$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0.$ <\/span><\/span> "}]}],"id_ques":161},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["<"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $Q=\\dfrac{x+2}{{{x}^{2}}+x+1}-\\dfrac{2}{x-1}$$-\\dfrac{2{{x}^{2}}+4}{1-{{x}^{3}}} $ <br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> So s\u00e1nh $Q\\,\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\,\\dfrac{1}{3}$ <br\/> ( \u0110i\u1ec1n d\u1ea5u > ; < ho\u1eb7c = v\u00e0o \u00f4 tr\u1ed1ng ) <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $Q=\\dfrac{x}{x^2+x+1}$ <br\/> Ta x\u00e9t hi\u1ec7u:<br\/> $\\begin{aligned} & Q-\\dfrac{1}{3} \\\\ & =\\dfrac{x}{x^2+x+1}-\\dfrac{1}{3} \\\\ & =\\dfrac{3x}{3\\left( {{x}^{2}}+x+1 \\right)}-\\dfrac{{{x}^{2}}+x+1}{3\\left( {{x}^{2}}+x+1 \\right)} \\\\ & =\\dfrac{3x-{{x}^{2}}-x-1}{3\\left( {{x}^{2}}+x+1 \\right)} \\\\ & =\\dfrac{-{{x}^{2}}+2x-1}{3\\left( {{x}^{2}}+x+1 \\right)} \\\\ & =\\dfrac{-{{\\left( x-1 \\right)}^{2}}}{3\\left( {{x}^{2}}+x+1 \\right)} \\\\ & Do\\,\\,\\left\\{ \\begin{aligned} & {{\\left( x-1 \\right)}^{2}} > 0\\,\\, \\\\ & {{x}^{2}}+x+1 > 0\\,\\, \\\\ \\end{aligned} \\right.\\forall x\\ne 1 \\\\ & \\Rightarrow \\dfrac{-{{\\left( x-1 \\right)}^{2}}}{3\\left( {{x}^{2}}+x+1 \\right)} < 0\\,\\,\\forall x\\ne 1 \\\\ & \\Rightarrow Q < \\dfrac{1}{3}\\,\\,\\,\\forall x\\ne 1 \\\\ \\end{aligned}$. <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $<$. <\/span><\/span> "}]}],"id_ques":162},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank_random","correct":[[["-1"],["1"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/4.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $N=\\dfrac{1}{x-1}-\\dfrac{1}{x+1}+\\dfrac{2}{{{x}^{2}}-1}$ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $N$ l\u00e0: $\\left\\{ \\begin{aligned} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $N$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x+1\\ne 0 \\\\ & x-1\\ne 0 \\\\ & x^2 - 1 \\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne -1 \\\\ & x\\ne 1 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$ v\u00e0 $1.$ <\/span><\/span> "}]}],"id_ques":163},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["4"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/4.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $N=\\dfrac{1}{x-1}-\\dfrac{1}{x+1}+\\dfrac{2}{{{x}^{2}}-1}$ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $N$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $N = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{{{x}^{2}}-1}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n: $x\\ne \\{-1;1\\}$ <br\/> Ta c\u00f3: <br\/> $ N=\\dfrac{1}{x-1}-\\dfrac{1}{x+1}+\\dfrac{2}{{{x}^{2}}-1} $<br\/>$=\\dfrac{x+1}{\\left( x+1 \\right)\\left( x-1 \\right)}$$-\\dfrac{x-1}{\\left( x+1 \\right)\\left( x-1 \\right)}$$+\\dfrac{2}{\\left( x+1 \\right)\\left( x-1 \\right)} $<br\/>$ =\\dfrac{x+1-x+1+2}{\\left( x+1 \\right)\\left( x-1 \\right)} $<br\/>$=\\dfrac{4}{{{x}^{2}}-1}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4$. <\/span><\/span> "}]}],"id_ques":164},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["4"],["15"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/4.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $N=\\dfrac{1}{x-1}-\\dfrac{1}{x+1}+\\dfrac{2}{{{x}^{2}}-1}$ <br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $x=4$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $N$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":" Thay $x=4$ v\u00e0o bi\u1ec3u th\u1ee9c $N$ \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $N=\\dfrac{4}{{{x}^{2}}-1}$ <br\/> Thay $x=4$ v\u00e0o bi\u1ec3u th\u1ee9c $N$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $N=\\dfrac{4}{{{x}^{2}}-1}$$=\\dfrac{4}{4^2-1}=\\dfrac{4}{15}$<\/span> "}]}],"id_ques":165},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/4.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $N=\\dfrac{1}{x-1}-\\dfrac{1}{x+1}+\\dfrac{2}{{{x}^{2}}-1}$ <br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> Ta lu\u00f4n c\u00f3 $N > 0$ v\u1edbi m\u1ecdi $x\\ne \\{-1;1\\}$ <\/span> ","select":["\u0110\u00fang","Sai"],"hint":" Ta nh\u1eadn \u0111\u1ecbnh gi\u00e1 tr\u1ecb c\u1ee7a t\u1eed th\u1ee9c v\u00e0 m\u1eabu th\u1ee9c c\u1ee7a bi\u1ec3u th\u1ee9c $N$ \u0111\u00e3 r\u00fat g\u1ecdn. ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $N=\\dfrac{4}{{{x}^{2}}-1}$ <br\/> Ta nh\u1eadn th\u1ea5y t\u1eed th\u1ee9c c\u1ee7a $N$ l\u00e0 $4 > 0$ <br\/> M\u1eabu th\u1ee9c c\u1ee7a $N$ l\u00e0 $x^2-1$ ch\u01b0a ch\u1eafc l\u00e0 s\u1ed1 d\u01b0\u01a1ng v\u1edbi m\u1ecdi $x\\ne \\{-1;1\\}$.<br\/> Do \u0111\u00f3 k\u1ebft lu\u1eadn $N > 0$ v\u1edbi m\u1ecdi $x\\ne \\{-1;1\\}$ l\u00e0 <b> sai<\/b>. <span><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span>","column":2}]}],"id_ques":166},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["2"],["-3","3"],["3","-3"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/3.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $M=\\dfrac{{{x}^{2}}-4}{9-{{y}^{2}}}:\\dfrac{x-2}{y+3}$$-\\dfrac{2}{3-y} $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $M$ l\u00e0: $\\left\\{ \\begin{align} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & y\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & y\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$<\/span> ","hint":" \u0110\u1ec3 ph\u00e2n th\u1ee9c c\u00f3 ngh\u0129a th\u00ec m\u1eabu th\u1ee9c ph\u1ea3i kh\u00e1c 0, ph\u00e2n th\u1ee9c chia kh\u00e1c 0. ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $M$ l\u00e0:<br\/> $\\left\\{ \\begin{align} & 9-{{y}^{2}}\\ne 0 \\\\ & y+3\\ne 0 \\\\ & x-2\\ne 0 \\\\ & 3-y\\ne 0 \\\\ \\end{align} \\right.$$\\Rightarrow \\left\\{ \\begin{align} & y+3\\ne 0 \\\\ & y-3\\ne 0 \\\\ & x-2\\ne 0 \\\\ \\end{align} \\right.$$\\Rightarrow \\left\\{ \\begin{align} & y\\ne -3 \\\\ & y\\ne 3 \\\\ & x\\ne 2 \\\\ \\end{align} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2; 3$ v\u00e0 $-3.$ <\/span><\/span> "}]}],"id_ques":167},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["x"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/3.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $M=\\dfrac{{{x}^{2}}-4}{9-{{y}^{2}}}:\\dfrac{x-2}{y+3}$$-\\dfrac{2}{3-y} $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $M$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $M = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{3-y}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $M$ l\u00e0: $x\\ne 2; y\\ne \\{-3;3\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & M=\\dfrac{{{x}^{2}}-4}{9-{{y}^{2}}}:\\dfrac{x-2}{y+3}-\\dfrac{2}{3-y} \\\\ & =\\dfrac{{{x}^{2}}-4}{9-{{y}^{2}}}\\cdot \\dfrac{y+3}{x-2}-\\dfrac{2}{3-y} \\\\ & =\\dfrac{\\left( x+2 \\right)\\left( x-2 \\right)}{\\left( 3+y \\right)\\left( 3-y \\right)}\\cdot \\dfrac{y+3}{x-2}-\\dfrac{2}{3-y} \\\\ & =\\dfrac{x+2}{3-y}-\\dfrac{2}{3-y} \\\\ & =\\dfrac{x+2-2}{3-y} \\\\ & =\\dfrac{x}{3-y} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $x$. <\/span><\/span> "}]}],"id_ques":168},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["2016"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/3.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $M=\\dfrac{{{x}^{2}}-4}{9-{{y}^{2}}}:\\dfrac{x-2}{y+3}$$-\\dfrac{2}{3-y} $ <br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $x=2016;y=2$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $M$ l\u00e0 _input_ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2, ta c\u00f3: $M=\\dfrac{x}{3-y}$ <br\/> Thay $x=2016;y=2$ v\u00e0o bi\u1ec3u th\u1ee9c $M$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $M=\\dfrac{x}{3-y}$$=\\dfrac{2016}{3-2}=2016$<\/span>"}]}],"id_ques":169},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv1/img\/3.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $M=\\dfrac{{{x}^{2}}-4}{9-{{y}^{2}}}:\\dfrac{x-2}{y+3}$$-\\dfrac{2}{3-y} $ <br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x$ \u0111\u1ec3 $M=0$<br\/> \u0110\u00e1p \u00e1n: $x =$ _input_ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2, ta c\u00f3: $M=\\dfrac{x}{3-y}$ <br\/> $M = 0$ $\\Leftrightarrow\\dfrac{x}{3-y}=0 \\Rightarrow x=0 $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0.$ <\/span><\/span> "}]}],"id_ques":170}],"lesson":{"save":0,"level":1}}