{"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","correct":[[["4a"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{2+a}{2-a}-\\dfrac{4{{a}^{2}}}{{{a}^{2}}-4}-\\dfrac{2-a}{2+a} \\right)$$\\cdot \\dfrac{{{a}^{2}}-2a}{2{{a}^{2}}-a}$ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $C$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $C = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{1-2a}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $C$ l\u00e0: $a\\ne \\{-2;0;\\dfrac{1}{2};2\\}$ <br\/> $\\begin{align} & C=\\left( \\dfrac{2+a}{2-a}-\\dfrac{4{{a}^{2}}}{{{a}^{2}}-4}-\\dfrac{2-a}{2+a} \\right)\\cdot \\dfrac{{{a}^{2}}-2a}{2{{a}^{2}}-a} \\\\ & =\\left[ \\dfrac{{{\\left( 2+a \\right)}^{2}}}{4-{{a}^{2}}}+\\dfrac{4{{a}^{2}}}{4-{{a}^{2}}}-\\dfrac{{{\\left( 2-a \\right)}^{2}}}{4-{{a}^{2}}} \\right]\\cdot \\dfrac{a\\left( a-2 \\right)}{a\\left( 2a-1 \\right)} \\\\ & =\\dfrac{{{\\left( 2+a \\right)}^{2}}+4{{a}^{2}}-{{\\left( 2-a \\right)}^{2}}}{4-{{a}^{2}}}\\cdot \\dfrac{a-2}{2a-1} \\\\ & =\\dfrac{4+4a+{{a}^{2}}+4{{a}^{2}}-4+4a-{{a}^{2}}}{4-{{a}^{2}}}\\cdot \\dfrac{a-2}{2a-1} \\\\ & =\\dfrac{4{{a}^{2}}+8a}{\\left( 2+a \\right)\\left( 2-a \\right)}\\cdot \\dfrac{a-2}{2a-1} \\\\ & =-\\dfrac{4a\\left( a+2 \\right)}{\\left( 2+a \\right)}\\cdot \\dfrac{1}{2a-1} \\\\ & =-\\dfrac{4a}{2a-1} \\\\ & =\\dfrac{4a}{1-2a} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4a$. <\/span><\/span> "}]}],"id_ques":191},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{2+a}{2-a}-\\dfrac{4{{a}^{2}}}{{{a}^{2}}-4}-\\dfrac{2-a}{2+a} \\right)$$\\cdot \\dfrac{{{a}^{2}}-2a}{2{{a}^{2}}-a}$<br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> T\u00ecm $a$ nguy\u00ean \u0111\u1ec3 $C\\in \\mathbb{Z}$ <br\/> \u0110\u00e1p \u00e1n: $a=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $C=\\dfrac{4a}{1-2a}$<br\/> $\\Rightarrow C=-\\dfrac{4a}{2a-1} $ <br\/> $=-\\dfrac{2\\left( 2a-1 \\right)+2}{2a-1} $ <br\/> $ =-\\left( 2+\\dfrac{2}{2a-1} \\right) $ <br\/> \u0110\u1ec3 $C\\in \\mathbb{Z}$ th\u00ec $\\dfrac{2}{2a-1}\\in \\mathbb{Z}$ <br\/> Khi \u0111\u00f3 $2a-1\\in \u01af\\left( 2 \\right)$ , m\u00e0 $\u01af\\left( 2 \\right)=\\left\\{ -2;-1;1;2 \\right\\}$ <br\/> Ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$2a-1$<\/th> <th>$-2$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <th>$2$<\/th> <\/tr> <tr> <td>$a$<\/td> <td>$-\\dfrac{1}{2}$ (lo\u1ea1i)<\/td> <td>0 (lo\u1ea1i)<\/td> <td>$1$<\/td> <td>$\\dfrac{3}{2}$ (lo\u1ea1i)<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1.$ <\/span><\/span> "}]}],"id_ques":192},{"time":24,"part":[{"title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\left[\\begin{array}{l}{x= \\dfrac{-32}{15}} \\\\ {x = \\dfrac{-8}{3}}\\end{array}\\right.$","B. $\\left[\\begin{array}{l}{x= \\dfrac{-28}{15}} \\\\ {x = \\dfrac{8}{3}}\\end{array}\\right.$","C. $\\left[\\begin{array}{l}{x= \\dfrac{-22}{15}} \\\\ {x = \\dfrac{-8}{3}}\\end{array}\\right.$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{2+a}{2-a}-\\dfrac{4{{a}^{2}}}{{{a}^{2}}-4}-\\dfrac{2-a}{2+a} \\right)$$\\cdot \\dfrac{{{a}^{2}}-2a}{2{{a}^{2}}-a}$<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> N\u1ebfu $|a-5|=3$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $C$ l\u00e0 ? <\/span> ","explain":"<span class='basic_left'> Theo b\u00e0i: <br\/> $\\begin{aligned} & \\left| a-5 \\right|=3 \\\\ & \\Rightarrow \\left[ \\begin{aligned} & a-5=3\\,\\,\\,\\left( a\\ge 5 \\right) \\\\ & a-5=-3\\,\\,\\,\\left( a<5 \\right) \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow \\left[ \\begin{aligned} & a=8\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ & a=2\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> Theo c\u00e2u 1 tr\u00ean, Ta c\u00f3: <br\/> $C=\\dfrac{4a}{1-2a}$ <br\/> Thay $a=8$ v\u00e0o C, ta \u0111\u01b0\u1ee3c:<br\/> $ \\dfrac{4a}{1-2a}=\\dfrac{4.8}{1-2.8}=\\dfrac{-32}{15} $ <br\/> Thay $a=2$ v\u00e0o C, ta \u0111\u01b0\u1ee3c:<br\/> $ \\dfrac{4a}{1-2a}=\\dfrac{4.2}{1-2.2}=\\dfrac{-8}{3} $<\/span> "}]}],"id_ques":193},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["1"],["10"]]],"list":[{"point":10,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/10.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{2+a}{2-a}-\\dfrac{4{{a}^{2}}}{{{a}^{2}}-4}-\\dfrac{2-a}{2+a} \\right)$$\\cdot \\dfrac{{{a}^{2}}-2a}{2{{a}^{2}}-a}$<br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> \u0110\u1ec3 $C = \\dfrac{1}{2}$ th\u00ec $a =$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $C=\\dfrac{4a}{1-2a}$ <br\/> \u0110\u1ec3 $C=\\dfrac{1}{2}$ th\u00ec <br\/> $\\begin{align} & \\dfrac{4a}{1-2a}=\\frac{1}{2} \\\\ &\\Leftrightarrow 4a.2=\\left( 1-2a \\right).1 \\\\ & \\Leftrightarrow 8a=1-2a \\\\ & \\Leftrightarrow 10a=1 \\\\ & \\Leftrightarrow a=\\dfrac{1}{10}\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ \\end{align}$<\/span> "}]}],"id_ques":194},{"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"],["0"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ A=\\dfrac{1}{{{x}^{2}}+x+1}-\\dfrac{1}{x-{{x}^{2}}}$$-\\dfrac{{{x}^{2}}+2x}{{{x}^{3}}-1} $ <br\/> <br\/> <br\/> <b> C\u00e2u 1: <\/b> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $A$ l\u00e0: $\\left\\{ \\begin{aligned} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{aligned} \\right.$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c $A$ l\u00e0:<br\/> $\\left\\{ \\begin{aligned} & x^3-1\\ne 0 \\\\ & x-x^2\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> $\\Rightarrow \\left\\{ \\begin{aligned} & (x-1)(x^2+x+1)\\ne 0 \\\\ & x(1-x)\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> $\\Rightarrow \\left\\{ \\begin{aligned} & x-1\\ne 0 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> $\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 1 \\\\ & x\\ne 0 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0; 1.$ <\/span><\/span> "}]}],"id_ques":195},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["-1"],["x"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ A=\\dfrac{1}{{{x}^{2}}+x+1}-\\dfrac{1}{x-{{x}^{2}}}$$-\\dfrac{{{x}^{2}}+2x}{{{x}^{3}}-1} $ <br\/> <br\/> <br\/> <b> C\u00e2u 2: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $A = \\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 1 tr\u00ean, \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c A l\u00e0: $x\\ne \\{0;1\\}$ <br\/>Ta c\u00f3: <br\/> $\\begin{align} & A=\\dfrac{1}{{{x}^{2}}+x+1}-\\dfrac{1}{x-{{x}^{2}}}-\\dfrac{{{x}^{2}}+2x}{{{x}^{3}}-1} \\\\ & =\\dfrac{x-1}{{{x}^{3}}-1}-\\dfrac{1}{x\\left( 1-x \\right)}-\\dfrac{{{x}^{2}}+2x}{{{x}^{3}}-1} \\\\ & =\\dfrac{x-1}{{{x}^{3}}-1}-\\dfrac{{{x}^{2}}+2x}{{{x}^{3}}-1}+\\dfrac{1}{x\\left( x-1 \\right)} \\\\ & =\\dfrac{x-1-{{x}^{2}}-2x}{{{x}^{3}}-1}+\\dfrac{1}{x\\left( x-1 \\right)} \\\\ & =\\dfrac{-{{x}^{2}}-x-1}{{{x}^{3}}-1}+\\dfrac{1}{x\\left( x-1 \\right)} \\\\ & =\\dfrac{-1}{x-1}+\\dfrac{1}{x\\left( x-1 \\right)} \\\\ & =\\dfrac{-x+1}{x\\left( x-1 \\right)} \\\\ & =\\dfrac{-1}{x} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$ v\u00e0 $x$. <\/span><\/span> "}]}],"id_ques":196},{"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":[[["-8"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $A=\\dfrac{1}{{{x}^{2}}+x+1}-\\dfrac{1}{x-{{x}^{2}}}$$-\\dfrac{{{x}^{2}}+2x}{{{x}^{3}}-1} $<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $x=\\dfrac{1}{8}$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $A$ l\u00e0 _input_ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $A=\\dfrac{-1}{x}$ <br\/> Thay $x=\\dfrac{1}{8}$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn tr\u00ean, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{-1}{x}=\\dfrac{-1}{\\dfrac{1}{8}}=-8$<\/span> "}]}],"id_ques":197},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank_random","correct":[[["-2017"]]],"list":[{"point":10,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/8.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ A=\\dfrac{1}{{{x}^{2}}+x+1}-\\dfrac{1}{x-{{x}^{2}}}$$-\\dfrac{{{x}^{2}}+2x}{{{x}^{3}}-1} $<br\/> <br\/> <br\/> <b> C\u00e2u 4: <\/b> T\u00ecm $x$ \u0111\u1ec3 $A=\\dfrac{1}{2017}$ <br\/> <br\/> \u0110\u00e1p \u00e1n: $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $A=\\dfrac{-1}{x}$ <br\/> \u0110\u1ec3 $A = \\dfrac{1}{2017}$ th\u00ec <br\/> $\\begin{aligned} & \\dfrac{-1}{x}=\\dfrac{1}{2017} \\\\ &\\Leftrightarrow \\dfrac{1}{-x}=\\dfrac{1}{2017} \\\\ & \\Leftrightarrow -x=2017 \\\\ & \\Leftrightarrow x=-2017 \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2017$. <\/span><\/span> "}]}],"id_ques":198},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-2"],["-4"]]],"list":[{"point":10,"width":30,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/4.jpg' \/><\/center> <span class='basic_left'>T\u00ecm $x$ nguy\u00ean \u0111\u1ec3 $M=\\dfrac{x+4}{x+3} \\in \\mathbb{Z}$ <br\/> <br\/> <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x\\in \\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span> ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne 3$ <br\/> $\\begin{align} & M=\\dfrac{x+4}{x+3} \\\\ & =\\dfrac{x+3+1}{x+3} \\\\ & =1+\\dfrac{1}{x+3} \\\\ \\end{align}$ <br\/> \u0110\u1ec3 $M\\in \\mathbb{Z}$ th\u00ec $\\dfrac{1}{x+3}\\in \\mathbb{Z}$ <br\/> Khi \u0111\u00f3 $x+3\\in \u01af\\left( 1 \\right)$ , m\u00e0 $\u01af\\left( 1 \\right)=\\left\\{ -1;1 \\right\\}$ <br\/> Ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$x+3$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <\/tr> <tr> <td>$x$<\/td> <td>$-4$<\/td> <td>$-2$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-4;-2.$ <\/span><\/span> "}]}],"id_ques":199},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-2"],["2"],["4"],["0"]]],"list":[{"point":10,"width":30,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/2.jpg' \/><\/center> <span class='basic_left'>T\u00ecm $x$ nguy\u00ean \u0111\u1ec3 $Q=\\dfrac{-x-2}{1-x} \\in \\mathbb{Z}$ <br\/> <br\/> <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x\\in \\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span> ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne 1$ <br\/> $\\begin{align} & Q=\\dfrac{-x-2}{1-x} \\\\ & =\\dfrac{x+2}{x-1} \\\\ & =\\dfrac{x-1+3}{x-1} \\\\ & =\\dfrac{x-1}{x-1}+\\dfrac{3}{x-1} \\\\ & =1+\\dfrac{3}{x-1} \\\\ \\end{align}$ <br\/> \u0110\u1ec3 $Q\\in \\mathbb{Z}$ th\u00ec $\\dfrac{3}{x-1}\\in \\mathbb{Z}$ <br\/> Khi \u0111\u00f3 $x-1\\in \u01af\\left( 3 \\right)$ , m\u00e0 $\u01af\\left( 3 \\right)=\\left\\{ -3;-1;1;3 \\right\\}$ <br\/> Ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$x-1$<\/th> <th>$-3$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <th>$3$<\/th> <\/tr> <tr> <td>$x$<\/td> <td>$-2$<\/td> <td>$0$<\/td> <td>$2$<\/td> <td>$4$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2; 0; 2; 4.$ <\/span><\/span> "}]}],"id_ques":200},{"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"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/8.jpg' \/><\/center> <span class='basic_left'> Cho bi\u1ec3u th\u1ee9c $ Q=\\left( \\dfrac{a+1}{a-1}-\\dfrac{a-1}{a+1} \\right)$$:\\dfrac{2a}{5a-5} $<br\/> <br\/> <br\/> <b> C\u00e2u 3: <\/b> V\u1edbi $a=9$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $Q$ l\u00e0 _input_ <\/span> ","explain":"<span class='basic_left'> Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $Q=\\dfrac{10}{a+1}$ <br\/> Thay $a=9$ v\u00e0o bi\u1ec3u th\u1ee9c $Q$ \u0111\u00e3 r\u00fat g\u1ecdn tr\u00ean, ta \u0111\u01b0\u1ee3c:<br\/> $Q =\\dfrac{10}{9+1}$$=1$<\/span> "}]}],"id_ques":205},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"],["2"]]],"list":[{"point":5,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/3.jpg' \/><\/center> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x,$ th\u1ecfa m\u00e3n bi\u1ec3u th\u1ee9c $P=\\dfrac{2x-3}{1-\\dfrac{3}{x+2}}$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $0.$<br\/> \u0110\u00e1p \u00e1n: $x =$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":" R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $P$ sau \u0111\u00f3 cho $P=0$ \u0111\u1ec3 t\u00ecm $x.$ ","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 c\u00f3 ngh\u0129a v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $P.$ <br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho bi\u1ec3u th\u1ee9c $P$ \u0111\u00e3 r\u00fat g\u1ecdn b\u1eb1ng $0$ \u0111\u1ec3 t\u00ecm $x.$ <br\/> <b> B\u01b0\u1edbc 3:<\/b> So s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 k\u1ebft lu\u1eadn $x$ t\u00ecm \u0111\u01b0\u1ee3c. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{-2; 1\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & P=\\dfrac{2x-3}{1-\\dfrac{3}{x+2}} \\\\ & =\\dfrac{2x-3}{\\dfrac{x+2-3}{x+2}} \\\\ & =\\dfrac{2x-3}{\\dfrac{x-1}{x+2}} \\\\ & =\\dfrac{\\left( 2x-3 \\right)\\left( x+2 \\right)}{x-1} \\\\ \\end{align}$ <br\/> $P = 0 \\Leftrightarrow$ $ \\dfrac{\\left( 2x-3 \\right)\\left( x+2 \\right)}{x-1}=0 $ <br\/> $ \\Leftrightarrow (2x-3)(x+2)=0 $ <br\/> $\\Leftrightarrow \\left[ \\begin{aligned} & 2x-3=0 \\\\ & x+2=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{3}{2}\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ & x=-2\\,\\,\\left( \\text{lo\u1ea1i} \\right) \\\\ \\end{aligned} \\right.$<\/span> "}]}],"id_ques":208},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-5"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/2.jpg' \/><\/center> T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $x,$ th\u1ecfa m\u00e3n bi\u1ec3u th\u1ee9c $Q=\\dfrac{{{x}^{2}}-25}{x+\\dfrac{1}{x-5}}$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $0.$<br\/> \u0110\u00e1p \u00e1n: $x =$ _input_ ","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 c\u00f3 ngh\u0129a c\u1ee7a $Q$ v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $Q.$<br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho bi\u1ec3u th\u1ee9c $Q$ \u0111\u00e3 r\u00fat g\u1ecdn b\u1eb1ng $0$ \u0111\u1ec3 t\u00ecm $x.$ <br\/> <b> B\u01b0\u1edbc 3:<\/b> So s\u00e1nh v\u1edbi \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n \u0111\u1ec3 k\u1ebft lu\u1eadn $x$ t\u00ecm \u0111\u01b0\u1ee3c. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne 5$ v\u00e0 $x+\\dfrac{1}{x-5}\\ne 0$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & Q=\\dfrac{{{x}^{2}}-25}{x+\\dfrac{1}{x-5}} \\\\ & =\\dfrac{\\left( x+5 \\right)\\left( x-5 \\right)}{\\dfrac{x\\left( x-5 \\right)+1}{x-5}} \\\\ & =\\dfrac{\\left( x+5 \\right)\\left( x-5\\right)}{\\dfrac{{{x}^{2}}-5x+1}{x-5}} \\\\ & =\\dfrac{\\left( x+5 \\right){{\\left( x-5 \\right)}^{2}}}{{{x}^{2}}-5x+1} \\\\ \\end{align}$ <br\/> $Q = 0 \\Leftrightarrow$ $\\dfrac{\\left( x+5 \\right){{\\left( x-5 \\right)}^{2}}}{{{x}^{2}}-5x+1}=0 $ <br\/> $ \\Leftrightarrow (x+5)(x-5)^2=0 $ <br\/> $\\Leftrightarrow \\left[ \\begin{aligned} & x+5=0 \\\\ & x-5=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=-5\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ & x=5\\,\\,\\left( \\text{lo\u1ea1i} \\right) \\\\ \\end{aligned} \\right.$ <\/span> "}]}],"id_ques":209},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"],["5"]]],"list":[{"point":5,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/16.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $ P=\\left( 1+\\dfrac{3}{x+4} \\right)$$:\\left( x-5+\\dfrac{18}{x+4} \\right) $ t\u1ea1i $x = -3$ l\u00e0: <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x = -3$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'> Gi\u1ea3i <\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\neq \\{- 4;-1; 2\\}$ <br\/> Ta c\u00f3: <br\/> $ P=\\left( 1+\\dfrac{3}{x+4} \\right)$$:\\left( x-5+\\dfrac{18}{x+4} \\right) $ <br\/> $ =\\left( \\dfrac{x+4}{x+4}+\\dfrac{3}{x+4} \\right)$$:\\left[ \\dfrac{\\left( x-5 \\right)\\left( x+4 \\right)}{x+4}+\\dfrac{18}{x+4} \\right] $ <br\/> $ =\\dfrac{x+4+3}{x+4}$$:\\dfrac{\\left( x-5 \\right)\\left( x+4 \\right)+18}{x+4} $ <br\/> $ =\\dfrac{x+7}{x+4}$$:\\dfrac{{{x}^{2}}-x-20+18}{x+4} $ <br\/> $ =\\dfrac{x+7}{x+4}:\\dfrac{{{x}^{2}}-x-2}{x+4} $ <br\/> $ =\\dfrac{x+7}{x+4}\\cdot \\dfrac{x+4}{{{x}^{2}}-x-2} $ <br\/> $ =\\dfrac{x+7}{{{x}^{2}}+x-2x-2} $<br\/>$ =\\dfrac{x+7}{x\\left( x+1 \\right)-2\\left( x+1 \\right)} $ <br\/> $ =\\dfrac{x+7}{\\left( x+1 \\right)\\left( x-2 \\right)} $ <br\/> Thay $x = -3$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $P =\\dfrac{-3+7}{\\left( -3+1 \\right)\\left( -3-2 \\right)}=\\dfrac{2}{5}$ <\/span> "}]}],"id_ques":210},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["6"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/13.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $N=\\dfrac{{{x}^{2}}}{x-2}\\cdot \\left( \\dfrac{{{x}^{2}}+4}{x}-4 \\right)+3$ t\u1ea1i $x = 3$ l\u00e0: _input_ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x = 3$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\neq \\{2; 0\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & N=\\dfrac{{{x}^{2}}}{x-2}\\cdot \\left( \\dfrac{{{x}^{2}}+4}{x}-4 \\right)+3 \\\\ & =\\dfrac{{{x}^{2}}}{x-2}\\cdot \\left( \\dfrac{{{x}^{2}}+4}{x}-\\dfrac{4x}{x} \\right)+3 \\\\ & =\\dfrac{{{x}^{2}}}{x-2}\\cdot \\dfrac{{{x}^{2}}+4-4x}{x}+3 \\\\ & =\\dfrac{{{x}^{2}}}{x-2}\\cdot \\dfrac{{{\\left( x-2 \\right)}^{2}}}{x}+3 \\\\ & =x\\left( x-2 \\right)+3 \\\\ & ={{x}^{2}}-2x+3 \\\\ \\end{align}$ <br\/> Thay $x = 3$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $N =3^2-2.3+3=6$<\/span> "}]}],"id_ques":211},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":5,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai15/lv3/img\/12.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $ P=\\left( 2+\\dfrac{5}{{{x}^{2}}-1}+\\dfrac{x}{1-x}+\\dfrac{x}{2x+2} \\right)$$:\\dfrac{6}{{{x}^{2}}-1} $ t\u1ea1i $x = 2$ l\u00e0: _input_ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Thay $x = 2$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ednh. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\neq \\pm 1$ <br\/> Ta c\u00f3: <br\/> $ P=\\left( 2+\\dfrac{5}{{{x}^{2}}-1}+\\dfrac{x}{1-x}+\\dfrac{x}{2x+2} \\right)$$:\\dfrac{6}{{{x}^{2}}-1} $<br\/>$ =\\left[ 2+\\dfrac{5}{\\left( x+1 \\right)\\left( x-1 \\right)}-\\dfrac{x}{x-1}+\\dfrac{x}{2\\left( x+1 \\right)} \\right]$$:\\dfrac{6}{{{x}^{2}}-1} $<br\/>$ =\\left[ \\dfrac{4\\left( {{x}^{2}}-1 \\right)}{2\\left( {{x}^{2}}-1 \\right)}+\\dfrac{10}{2\\left( {{x}^{2}}-1 \\right)}-\\dfrac{2x\\left( x+1 \\right)}{2\\left( {{x}^{2}}-1 \\right)}+\\dfrac{x\\left( x-1 \\right)}{2\\left( {{x}^{2}}-1 \\right)} \\right]$$:\\dfrac{6}{{{x}^{2}}-1} $<br\/>$ =\\dfrac{4\\left( {{x}^{2}}-1 \\right)+10-2x\\left( x+1 \\right)+x\\left( x-1 \\right)}{2\\left( {{x}^{2}}-1 \\right)}$$\\cdot \\dfrac{{{x}^{2}}-1}{6} $<br\/>$ =\\dfrac{4{{x}^{2}}-4+10-2{{x}^{2}}-2x+{{x}^{2}}-x}{2\\left( {{x}^{2}}-1 \\right)}$$\\cdot \\dfrac{{{x}^{2}}-1}{6} $<br\/>$ =\\dfrac{3{{x}^{2}}-3x+6}{2\\left( {{x}^{2}}-1 \\right)}$$\\cdot \\dfrac{{{x}^{2}}-1}{6} $<br\/>$ =\\dfrac{3\\left( {{x}^{2}}-x+2 \\right)}{12} $<br\/>$ =\\dfrac{{{x}^{2}}-x+2}{4}$ <br\/> Thay $x = 2$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $P =\\dfrac{{{2}^{2}}-2+2}{4}=1$<\/span> "}]}],"id_ques":212}],"lesson":{"save":0,"level":3}}