{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["-2"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/12.jpg' \/><\/center> T\u00ecm $a, b$ sao cho $\\dfrac{{{x}^{2}}+5}{\\left( x-2 \\right){{\\left( x+1 \\right)}^{2}}}$$=\\dfrac{a}{x-2}+\\dfrac{b}{{{\\left( x+1 \\right)}^{2}}}$ v\u1edbi $x\\ne \\{-1;2\\}$ <br\/> \u0110\u00e1p \u00e1n: $a =$ _input_ ; $b =$ _input_","hint":" Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c v\u00e0 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{a}{x-2}+\\dfrac{b}{{{\\left( x+1 \\right)}^{2}}}$<br\/> S\u1eed d\u1ee5ng \u0111\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 hai t\u1eed th\u1ee9c \u0111\u1ec3 t\u00ecm $a$ v\u00e0 $b$. ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &\\dfrac{a}{x-2}+\\dfrac{b}{{{\\left( x+1 \\right)}^{2}}}\\\\&=\\dfrac{a{{\\left( x+1 \\right)}^{2}}}{\\left( x-2 \\right){{\\left( x+1 \\right)}^{2}}}+\\dfrac{b\\left( x-2 \\right)}{\\left( x-2 \\right){{\\left( x+1 \\right)}^{2}}} \\\\ & =\\dfrac{a\\,{{x}^{2}}+2ax+a+bx-2b}{\\left( x-2 \\right){{\\left( x+1 \\right)}^{2}}} \\\\ & =\\dfrac{a\\,{{x}^{2}}+\\left( 2a+b \\right)x+a-2b}{\\left( x-2 \\right){{\\left( x+1 \\right)}^{2}}} \\\\ \\end{align}$<br\/> \u0110\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 hai t\u1eed th\u1ee9c : ${{x}^{2}}+5\\equiv a\\,{{x}^{2}}+\\left( 2a+b \\right)x+a-2b,$ ta \u0111\u01b0\u1ee3c:<br\/> $\\left\\{ \\begin{aligned} & a=1 \\\\ & 2a+b=0 \\\\ & a-2b=5 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & a=1 \\\\ & b=-2 \\\\ \\end{aligned} \\right.$<\/span> "}]}],"id_ques":51},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $a = \\dfrac{1}{2}; b = -1; c = \\dfrac{1}{2}$","B. $a = 1; b = -1; c = \\dfrac{1}{2}$","C. $a = 2; b = -1; c = \\dfrac{1}{2}$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/11.jpg' \/><\/center> T\u00ecm $a, b, c$ sao cho $\\dfrac{1}{x\\left( x+1 \\right)\\left( x+2 \\right)}$$=\\dfrac{a}{x}+\\dfrac{b}{x+1}+\\dfrac{c}{x+2}$ v\u1edbi $x\\ne \\{-2;-1;1\\}$","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $ \\dfrac{a}{x}+\\dfrac{b}{x+1}+\\dfrac{c}{x+2}$<br\/> $=\\dfrac{a\\left( x+1 \\right)\\left( x+2 \\right)}{x\\left( x+1 \\right)\\left( x+2 \\right)}+\\dfrac{bx\\left( x+2 \\right)}{x\\left( x+1 \\right)\\left( x+2 \\right)}$$+\\dfrac{cx\\left( x+1 \\right)}{x\\left( x+1 \\right)\\left( x+2 \\right)} $<br\/> $ =\\dfrac{a\\left( {{x}^{2}}+3x+2 \\right)+b\\left( {{x}^{2}}+2x \\right)+c\\left( {{x}^{2}}+x \\right)}{x\\left( x+1 \\right)\\left( x+2 \\right)} $<br\/> $=\\dfrac{a{{x}^{2}}+3ax+2a+b{{x}^{2}}+2bx+c{{x}^{2}}+cx}{x\\left( x+1 \\right)\\left( x+2 \\right)} $<br\/> $ =\\dfrac{\\left( a+b+c \\right){{x}^{2}}+\\left( 3a+2b+c \\right)x+2a}{x\\left( x+1 \\right)\\left( x+2 \\right)} $<br\/> \u0110\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1 hai t\u1eed th\u1ee9c:<br\/> $1\\equiv \\left( a+b+c \\right){{x}^{2}}$$+\\left( 3a+2b+c \\right)x+2a,$ ta \u0111\u01b0\u1ee3c:<br\/> $\\left\\{ \\begin{aligned} & a+b+c=0 \\\\ & 3a+2b+c=0 \\\\ & 2a=1 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & a=\\dfrac{1}{2} \\\\ & b+c=\\dfrac{-1}{2} \\\\ & 2b+c=-\\dfrac{3}{2} \\\\ \\end{aligned} \\right.$$\\Rightarrow \\left\\{ \\begin{aligned} & a=\\dfrac{1}{2} \\\\ & b=-1 \\\\ & c=\\dfrac{1}{2} \\\\ \\end{aligned} \\right.$<\/span> "}]}],"id_ques":52},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/10.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $ \\dfrac{1}{x+2}+\\dfrac{2}{2-x}+\\dfrac{x}{{{x}^{2}}-4}$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn $x.$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/>$ \\dfrac{1}{x+2}+\\dfrac{2}{2-x}+\\dfrac{x}{{{x}^{2}}-4}$ <br\/> N\u1ebfu bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn kh\u00f4ng ch\u1ee9a bi\u1ebfn th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn v\u00e0 ng\u01b0\u1ee3c l\u1ea1i.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> Ta c\u00f3: <br\/> $ \\dfrac{1}{x+2}+\\dfrac{2}{2-x}+\\dfrac{x}{{{x}^{2}}-4}$<br\/> $=\\dfrac{1}{x+2}-\\dfrac{2}{x-2}+\\dfrac{x}{\\left( x+2 \\right)\\left( x-2 \\right)} $<br\/> $ =\\dfrac{1\\left( x-2 \\right)}{\\left( x+2 \\right)\\left( x-2 \\right)}-\\dfrac{2\\left( x+2 \\right)}{\\left( x+2 \\right)\\left( x-2 \\right)}$$+\\dfrac{x}{\\left( x+2 \\right)\\left( x-2 \\right)} $<br\/> $ =\\dfrac{x-2-2x-4+x}{\\left( x+2 \\right)\\left( x-2 \\right)} $<br\/> $ =\\dfrac{-6}{{{x}^{2}}-4} $ <br\/> V\u1eady gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn $x$. <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0: Sai.<\/span>","column":2}]}],"id_ques":53},{"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":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/9.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{x-y}{xy}+\\dfrac{y-z}{yz}+\\dfrac{z-x}{zx}$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn $x, y, z.$ ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $\\dfrac{x-y}{xy}+\\dfrac{y-z}{yz}+\\dfrac{z-x}{zx}$ <br\/> N\u1ebfu bi\u1ec3u th\u1ee9c \u0111\u00e3 r\u00fat g\u1ecdn kh\u00f4ng ch\u1ee9a bi\u1ebfn th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn v\u00e0 ng\u01b0\u1ee3c l\u1ea1i.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} &\\dfrac{x-y}{xy}+\\dfrac{y-z}{yz}+\\dfrac{z-x}{zx}\\\\&=\\dfrac{\\left( x-y \\right)z}{xyz}+\\dfrac{\\left( y-z \\right)x}{xyz}+\\dfrac{\\left( z-x \\right)y}{xyz} \\\\ & =\\dfrac{xz-yz+xy-xz+yz-xy}{xyz} \\\\ & =\\dfrac{0}{xyz} \\\\ & =0 \\\\ \\end{align}$ <br\/> V\u1eady gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn $x, y, z$. <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0: \u0110\u00fang.<\/span>","column":2}]}],"id_ques":54},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["7"]]],"list":[{"point":5,"width":40,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/8.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{3}{x-3}-\\dfrac{6x}{9-{{x}^{2}}}+\\dfrac{x}{x+3}$ t\u1ea1i $x=4$ l\u00e0 _input_ ","hint":"T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 bi\u1ec3u th\u1ee9c c\u00f3 ngh\u0129a.<br\/> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c sau \u0111\u00f3 thay $x=4$ v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u1ec3 t\u00ednh. ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne \\{-3;3\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} &\\dfrac{3}{x-3}-\\dfrac{6x}{9-{{x}^{2}}}+\\dfrac{x}{x+3}\\\\&=\\dfrac{3\\left( x+3 \\right)}{{{x}^{2}}-9}+\\dfrac{6x}{{{x}^{2}}-9}+\\dfrac{x\\left( x-3 \\right)}{{{x}^{2}}-9} \\\\ & =\\dfrac{3x+9+6x+{{x}^{2}}-3x}{{{x}^{2}}-9} \\\\ & =\\dfrac{{{x}^{2}}+6x+9}{\\left( x+3 \\right)\\left( x-3 \\right)} \\\\ & =\\dfrac{{{\\left( x+3 \\right)}^{2}}}{\\left( x+3 \\right)\\left( x-3 \\right)} \\\\ & =\\dfrac{x+3}{x-3} \\\\ \\end{align}$ <br\/> Thay $x=4$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{x+3}{x-3}=\\dfrac{4+3}{4-3}=7$ <\/span> "}]}],"id_ques":55},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["5"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/5.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{{{x}^{3}}+1}{{{x}^{2}}-4}-\\dfrac{4}{x+2}-x$ t\u1ea1i $x=7$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne \\{-2;2\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} &\\dfrac{{{x}^{3}}+1}{{{x}^{2}}-4}-\\dfrac{4}{x+2}-x\\\\&=\\dfrac{{{x}^{3}}+1}{{{x}^{2}}-4}-\\dfrac{4\\left( x-2 \\right)}{{{x}^{2}}-4}-\\dfrac{x\\left( {{x}^{2}}-4 \\right)}{{{x}^{2}}-4} \\\\ & =\\dfrac{{{x}^{3}}+1-4x+8-{{x}^{3}}+4x}{{{x}^{2}}-4} \\\\ & =\\dfrac{9}{{{x}^{2}}-4} \\\\ \\end{align}$ <br\/> Thay $x=7$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{9}{{{x}^{2}}-4}=\\dfrac{9}{7^2-4}=\\dfrac{9}{45}=\\dfrac{1}{5}$<\/span> "}]}],"id_ques":56},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["6"],["5"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/4.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $\\dfrac{3x+5}{{{x}^{2}}-5x}+\\dfrac{25-x}{25-5x}$ t\u1ea1i $x=-1$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne \\{0;5\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{3x+5}{{{x}^{2}}-5x}+\\dfrac{25-x}{25-5x}\\\\&=\\dfrac{3x+5}{x\\left( x-5 \\right)}+\\dfrac{25-x}{5\\left( 5-x \\right)} \\\\ & =\\dfrac{\\left( 3x+5 \\right)5}{5x\\left( x-5 \\right)}+\\dfrac{\\left( x-25 \\right)x}{5x\\left( x-5 \\right)} \\\\ & =\\dfrac{15x+25+{{x}^{2}}-25x}{5x\\left( x-5 \\right)} \\\\ & =\\dfrac{{{x}^{2}}-10x+25}{5x\\left( x-5 \\right)} \\\\ & =\\dfrac{{{\\left( x-5 \\right)}^{2}}}{5x\\left( x-5 \\right)} \\\\ & =\\dfrac{x-5}{5x} \\\\ \\end{align}$ <br\/> Thay $x=-1$ v\u00e0o bi\u1ec3u th\u1ee9c r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $\\dfrac{x-5}{5x}=\\dfrac{-1-5}{5.\\left( -1 \\right)}=\\dfrac{6}{5}$<\/span> "}]}],"id_ques":57},{"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/bai13/lv1/img\/3.jpg' \/><\/center> T\u00ecm $x,$ bi\u1ebft $\\dfrac{2x-1}{x+1}+\\dfrac{x+1}{x-1}=3$<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\u1ee7a $x$ \u0111\u1ec3 c\u00e1c ph\u00e2n th\u1ee9c c\u00f3 ngh\u0129a.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Chuy\u1ec3n c\u00e1c h\u1ea1ng t\u1eed sang v\u1ebf tr\u00e1i, \u0111\u1ec3 v\u1ebf ph\u1ea3i b\u1eb1ng 0.<br\/> <b> B\u01b0\u1edbc 3:<\/b> Th\u1ef1c hi\u1ec7n r\u00fat g\u1ecdn $\\dfrac{2x-1}{x+1}+\\dfrac{x+1}{x-1}-3.$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{-1;1\\}$<br\/> Ta c\u00f3: <br\/> $ \\dfrac{2x-1}{x+1}+\\dfrac{x+1}{x-1}=3 $<br\/> $\\Leftrightarrow \\dfrac{2x-1}{x+1}+\\dfrac{x+1}{x-1}-3=0 $<br\/> $\\Leftrightarrow \\dfrac{\\left( 2x-1 \\right)\\left( x-1 \\right)}{{{x}^{2}}-1}+\\dfrac{\\left( x+1 \\right)\\left( x+1 \\right)}{{{x}^{2}}-1}$$-\\dfrac{3\\left( {{x}^{2}}-1 \\right)}{{{x}^{2}}-1}=0 $<br\/> $\\Leftrightarrow \\dfrac{2{{x}^{2}}-3x+1+{{x}^{2}}+2x+1-3{{x}^{2}}+3}{{{x}^{2}}-1}=0 $<br\/> $\\Leftrightarrow \\dfrac{-x+5}{{{x}^{2}}-1}=0 $<br\/> $ \\Rightarrow -x+5=0 $<br\/> $\\Leftrightarrow x=5 $ (th\u1ecfa m\u00e3n)<\/span> "}]}],"id_ques":58},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"],["2"]]],"list":[{"point":5,"width":40,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/2.jpg' \/><\/center> T\u00ecm $x,$ bi\u1ebft $\\dfrac{7x-3}{x+6}+\\dfrac{3x+8}{x+6}=0$<br\/> \u0110\u00e1p \u00e1n: $x=$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x+6\\ne 0 \\Rightarrow x\\ne -6$.<br\/> Ta c\u00f3: <br\/> $\\begin{align} & \\dfrac{7x-3}{x+6}+\\dfrac{3x+8}{x+6}=0 \\\\ &\\Leftrightarrow \\dfrac{7x-3+3x+8}{x+6}=0 \\\\ & \\Leftrightarrow\\dfrac{10x+5}{x+6}=0 \\\\ &\\Rightarrow 10x+5=0 \\\\ &\\Leftrightarrow x=\\dfrac{-1}{2}\\,\\,\\,\\,\\left( th\u1ecfa \\, m\u00e3n \\right) \\\\ \\end{align}$<\/span> "}]}],"id_ques":59},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","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/bai13/lv1/img\/16.jpg' \/><\/center> T\u00ecm $x,$ bi\u1ebft $\\dfrac{1}{x^2-x+1}-x=0$<br\/> \u0110\u00e1p \u00e1n: $\\left[ \\begin{align} & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ ","hint":" Th\u1ef1c hi\u1ec7n ph\u00e9p tr\u1eeb \u1edf v\u1ebf tr\u00e1i $\\dfrac{1}{x^2-x+1}-x$, r\u00fat g\u1ecdn \u0111\u1ec3 t\u00ecm $x.$","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x^2-x+1\\ne 0$ v\u1edbi m\u1ecdi $x.$<br\/> Ta c\u00f3: <br\/> $\\begin{aligned} & \\dfrac{1}{{{x}^{2}}-x+1}-x=0 \\\\ &\\Leftrightarrow \\dfrac{1}{{{x}^{2}}-x+1}-\\dfrac{x\\left( {{x}^{2}}-x+1 \\right)}{{{x}^{2}}-x+1}=0 \\\\ &\\Leftrightarrow \\dfrac{1-x\\left( {{x}^{2}}-x+1 \\right)}{{{x}^{2}}-x+1}=0 \\\\ &\\Leftrightarrow \\dfrac{1-{{x}^{3}}+{{x}^{2}}-x}{{{x}^{2}}-x+1}=0 \\\\ &\\Leftrightarrow \\dfrac{-{{x}^{3}}+{{x}^{2}}-x+1}{{{x}^{2}}-x+1}=0 \\\\ &\\Leftrightarrow \\dfrac{{{x}^{2}}\\left( 1 - x \\right)+\\left( 1 - x \\right)}{{{x}^{2}}-x+1}=0 \\\\ &\\Leftrightarrow \\dfrac{\\left( 1 - x \\right)\\left( {{x}^{2}}+1 \\right)}{{{x}^{2}}-x+1}=0 \\\\ &\\Rightarrow \\left( 1 - x \\right)\\left( {{x}^{2}}+1 \\right)=0 \\\\ &\\Leftrightarrow \\left( 1 - x \\right)=0 \\\\ &\\Leftrightarrow \\left (x \\right) =1 \\\\ \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 1. <\/span><\/span> "}]}],"id_ques":60},{"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/bai13/lv1/img\/13.jpg' \/><\/center> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p tr\u1eeb $x-\\dfrac{x^3}{x^2-y^2}$ l\u00e0: ","select":["A. $\\dfrac{2x^3-xy^2}{x^2-y^2}$ ","B. $\\dfrac{xy^2}{x^2-y^2}$","C. $\\dfrac{-xy^2}{x^2-y^2}$","D. $\\dfrac{xy}{x^2-y^2}$"],"explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &x-\\dfrac{{{x}^{3}}}{{{x}^{2}}-{{y}^{2}}}\\\\&=\\dfrac{x\\left( {{x}^{2}}-{{y}^{2}} \\right)}{{{x}^{2}}-{{y}^{2}}}-\\dfrac{{{x}^{3}}}{{{x}^{2}}-{{y}^{2}}} \\\\ & =\\dfrac{x\\left( {{x}^{2}}-{{y}^{2}} \\right)-{{x}^{3}}}{{{x}^{2}}-{{y}^{2}}} \\\\ & =\\dfrac{{{x}^{3}}-x{{y}^{2}}-{{x}^{3}}}{{{x}^{2}}-{{y}^{2}}} \\\\ & =\\dfrac{-x{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span><\/span> ","column":2}]}],"id_ques":61},{"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/bai13/lv1/img\/12.jpg' \/><\/center> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p tr\u1eeb $\\dfrac{2x}{x-4}-\\dfrac{5x-2}{x^2-16}$ l\u00e0: ","select":["A. $\\dfrac{2x^2+3x+2}{x^2-16}$ ","B. $\\dfrac{7x-2}{x^2-16}$","C. $\\dfrac{-3x+2}{x^2-16}$","D. $\\dfrac{x^2+x-1}{x^2-16}$"],"explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\begin{align} &\\dfrac{2x}{x-4}-\\dfrac{5x-2}{{{x}^{2}}-16}\\\\&=\\dfrac{2x}{x-4}-\\dfrac{5x-2}{\\left( x+4 \\right)\\left( x-4 \\right)} \\\\ & =\\dfrac{2x\\left( x+4 \\right)}{\\left( x+4 \\right)\\left( x-4 \\right)}-\\dfrac{5x-2}{\\left( x+4 \\right)\\left( x-4 \\right)} \\\\ & =\\dfrac{2x\\left( x+4 \\right)-\\left( 5x-2 \\right)}{\\left( x+4 \\right)\\left( x-4 \\right)} \\\\ & =\\dfrac{2{{x}^{2}}+8x-5x+2}{\\left( x+4 \\right)\\left( x-4 \\right)} \\\\ & =\\dfrac{2{{x}^{2}}+3x+2}{\\left( x+4 \\right)\\left( x-4 \\right)} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span> ","column":2}]}],"id_ques":62},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p tr\u1eeb c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["2y"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/11.jpg' \/><\/center> $\\dfrac{2x+y}{x(y^2-x)}-\\dfrac{2x-y}{x(y^2-x)}$$=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{x(y^2-x)}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\dfrac{2x+y}{x(y^2-x)}-\\dfrac{2x-y}{x(y^2-x)}$ <br\/> $=\\dfrac{2x+y-2x+y}{x(y^2-x)}\\\\ =\\dfrac{2y}{x(y^2-x)}$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2y$. <\/span><\/span> "}]}],"id_ques":63},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p c\u1ed9ng c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["y"],["1+x","1+x"]],[["x+1","1+x"],["y"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/10.jpg' \/><\/center> $\\dfrac{xy}{x-y^2}+\\dfrac{y}{x-y^2}$$=\\dfrac{(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}).(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})}{x-y^2}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\dfrac{xy}{x-y^2}+\\dfrac{y}{x-y^2}$ <br\/> $=\\dfrac{xy+y}{x-y^2}$ <br\/> $=\\dfrac{y(x+1)}{x-y^2}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{y(x+1)}{x-y^2}$ <\/span><\/span> "}]}],"id_ques":64},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng trong ph\u00e9p c\u1ed9ng c\u00e1c ph\u00e2n th\u1ee9c sau","title_trans":"","temp":"fill_the_blank","correct":[[["2x"],["a"]]],"list":[{"point":5,"width":40,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai13/lv1/img\/9.jpg' \/><\/center> $\\dfrac{x+y}{a}+\\dfrac{x-y}{a}=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ ","explain":"<span class='basic_left'> Ta c\u00f3:<br\/> $\\dfrac{x+y}{a}+\\dfrac{x-y}{a}$$=\\dfrac{x+y+x-y}{a}=\\dfrac{2x}{a}$ <br\/> <span class='basic_pink'> K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh tr\u00ean l\u00e0 $\\dfrac{2x}{a}$ <\/span><\/span> "}]}],"id_ques":65},{"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/bai13/lv1/img\/8.jpg' \/><\/center> Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c $\\dfrac{x-1}{x^3-1}$; $\\dfrac{x}{x^2+x+1}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $\\dfrac{x-1}{x^3-1}$; $\\dfrac{x(x-1)}{x^3-1}$ ","B. $\\dfrac{x-1}{x^2+x+1}$; $\\dfrac{x}{x^2+x+1}$ ","C. $\\dfrac{x-1}{x^3-1}$; $\\dfrac{x}{x^3-1}$","D. $\\dfrac{x-1}{x^2+x+1}$; $\\dfrac{x}{x^2+x+1}$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> <b> Mu\u1ed1n quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c:<\/b> <br\/> + T\u00ecm m\u1eabu th\u1ee9c chung. <br\/> + X\u00e1c \u0111\u1ecbnh c\u00e1c nh\u00e2n t\u1eed ph\u1ee5: nh\u00e2n t\u1eed ph\u1ee5 l\u00e0 th\u01b0\u01a1ng c\u1ee7a m\u1eabu th\u1ee9c chung v\u1edbi t\u1eebng m\u1eabu th\u1ee9c. <br\/> + Nh\u00e2n t\u1eed v\u00e0 m\u1eabu c\u1ee7a m\u1ed7i ph\u00e2n th\u1ee9c v\u1edbi nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a n\u00f3. <br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/><span class='basic_left'> M\u1eabu th\u1ee9c chung: $(x-1)(x^2+x+1)$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $x^3-1$ l\u00e0: $1$. Do \u0111\u00f3:<br\/> $\\dfrac{x-1}{x^3-1}=\\dfrac{(x-1).1}{(x^3-1).1}=\\dfrac{x-1}{x^3-1}$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $x^2+x+1$ l\u00e0: $x-1$. Do \u0111\u00f3:<br\/> $\\dfrac{x}{x^2+x+1}=\\dfrac{x.(x-1)}{(x^2+x+1).(x-1)}$$=\\dfrac{x(x-1)}{x^3-1}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":66},{"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/bai13/lv1/img\/5.jpg' \/><\/center> Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c $\\dfrac{1}{x}$; $\\dfrac{2}{x^2}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0:","select":["A. $\\dfrac{1}{x^2}$; $\\dfrac{2}{x^2}$ ","B. $\\dfrac{x}{x^2}$; $\\dfrac{2}{x^2}$ ","C. $\\dfrac{x-1}{x^2}$; $\\dfrac{2}{x^2}$","D. $\\dfrac{x-1}{x^2-1}$; $\\dfrac{2}{x^2-1}$"],"explain":"<span class='basic_left'> M\u1eabu th\u1ee9c chung: $x^2$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $x$ l\u00e0: $x^2:x=x$. Do \u0111\u00f3:<br\/> $\\dfrac{1}{x}=\\dfrac{1.x}{x.x}=\\dfrac{x}{x^2}$ <br\/> Nh\u00e2n t\u1eed ph\u1ee5 c\u1ee7a m\u1eabu th\u1ee9c $x^2$ l\u00e0: $x^2:x^2=1$. Do \u0111\u00f3:<br\/> $\\dfrac{2}{x^2}=\\dfrac{2.1}{x^2.1}=\\dfrac{2}{x^2}$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":67},{"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/bai13/lv1/img\/4.jpg' \/><\/center> M\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c $\\dfrac{1}{x^3+1}$ v\u00e0 $\\dfrac{x}{x+1}$ l\u00e0:","select":["A. $x^2+x+1$ ","B. $x^3$ ","C. $x+1$","D. $x^3+1$"],"explain":"<span class='basic_left'> Ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed, ta c\u00f3:<br\/> $x^3+1=(x+1)(x^2-x+1)$ <br\/> V\u1eady m\u1eabu th\u1ee9c chung l\u00e0: $x^3+1$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":68},{"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/bai13/lv1/img\/3.jpg' \/><\/center> M\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c $\\dfrac{1}{x^2-x}$ v\u00e0 $\\dfrac{8}{x}$ l\u00e0:","select":["A. $x^2$ ","B. $x^2-x$ ","C. $x-1$","D. $x^3-x$"],"explain":"<span class='basic_left'> Ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed, ta c\u00f3:<br\/> $x^2-x=x(x-1)$ <br\/> V\u1eady m\u1eabu th\u1ee9c chung l\u00e0: $x(x-1)$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":69},{"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/bai13/lv1/img\/2.jpg' \/><\/center> M\u1eabu th\u1ee9c chung c\u1ee7a hai ph\u00e2n th\u1ee9c $\\dfrac{3}{x^2y^3}$ v\u00e0 $\\dfrac{7}{10x^5y^2}$ l\u00e0:","select":["A. $10x^5y^3$ ","B. $x^5y^3$ ","C. $10x^5y^2$","D. $x^5y^2$"],"explain":"<span class='basic_left'> $BCNN(1;10)=10$<br\/> S\u1ed1 m\u0169 cao nh\u1ea5t c\u1ee7a $x$ l\u00e0 $5;$ s\u1ed1 m\u0169 cao nh\u1ea5t c\u1ee7a $y$ l\u00e0 $3.$<br\/> V\u1eady m\u1eabu th\u1ee9c chung l\u00e0: $10x^5y^3$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":2}]}],"id_ques":70}],"lesson":{"save":0,"level":1}}