\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)<=>  \(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)

<=>\(\frac{8+x-8}{x-8}+\frac{11+x-11}{x-11}=\frac{9+x-9}{x-9}+\frac{10+x-10}{x-10}\)

<=>\(\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)

<=>\(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=>\(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

=>\(\orbr{\begin{cases}x=0\\\frac{1}{x-8}+\frac{1}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\end{cases}}\)

đến đoạn bạn giải tiếp nhé

\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)

\(\Leftrightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}\right)\)

\(\Leftrightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)

ta chuyển về vế trái được 

\(\Leftrightarrow\left(x+18\right)\left(\frac{1}{13}+\frac{1}{122}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

\(\Leftrightarrow x+2018=0\)(do cái còn lại khác 0)

\(\Leftrightarrow x=-2018\)

mình nghĩ đề cậu viết thiếu mình sửa rồi

Ta có:

\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)

\(\Rightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}+1\right)\)

\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)

\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}-\frac{x+18}{10}-\frac{x+18}{9}-\frac{x+18}{8}=0\)

\(\Rightarrow\left(x+18\right)\times\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

Vì \(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)

\(\Rightarrow x+18=0\)

\(\Rightarrow x=-18\)

Vậy phương trình có nghiệm là x = -18

xin lỗi vì tớ nhầm một chút ở dòng cuối cho mình xin lỗi nhé

a/ ĐKXĐ: \(x\ne\left\{8;9;10;11\right\}\)

\(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)

\(\Leftrightarrow\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)

\(\Leftrightarrow x\left(\frac{1}{x-8}-\frac{1}{x-9}+\frac{1}{x-11}-\frac{1}{x-10}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-9}-\frac{1}{x-8}=\frac{1}{x-11}-\frac{1}{x-10}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{1}{\left(x-9\right)\left(x-8\right)}=\frac{1}{\left(x-11\right)\left(x-10\right)}\)

\(\Leftrightarrow x^2-17x+72=x^2-21x+110\)

\(\Rightarrow x=\frac{19}{2}\)

b/ ĐK: \(x\ne\left\{3;4;5;6\right\}\)

\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-3}-\frac{1}{x-5}=\frac{1}{x-4}-\frac{1}{x-6}\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\frac{-2}{\left(x-3\right)\left(x-5\right)}=\frac{-2}{\left(x-4\right)\left(x-6\right)}\)

\(\Leftrightarrow x^2-8x+15=x^2-10x+24\)

\(\Rightarrow x=\frac{9}{2}\)

\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

\(-537x^2+5054x=-541x^2+5092x\)

\(-537x^2+5054x+541x^2-5092x=0\)

\(4x^2-38x=0\)

\(x\left(2x-19\right)=0\)

\(\orbr{\begin{cases}x=0\\2x=19\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=\frac{19}{2}\end{cases}}\)

ĐKXĐ \(x\ne8;x\ne11;x\ne9;x\ne10\)

\(\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)

\(\Leftrightarrow\left(\dfrac{8}{x-8}+1\right)+\left(\dfrac{11}{x-11}+1\right)=\left(\dfrac{9}{x-9}+1\right)+\left(\dfrac{10}{x-10}+1\right)\)

\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}=\dfrac{x}{x-9}+\dfrac{x}{x-10}\)

\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\)

\(\Leftrightarrow x=0\) hoặc \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)

1) x=0

2) \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)

\(\Leftrightarrow\dfrac{x-11+x-8}{\left(x-8\right)\left(x-11\right)}-\dfrac{x-10+x-9}{\left(x-9\right)\left(x-10\right)}=0\)

\(\Leftrightarrow\dfrac{2x-19}{\left(x-8\right)\left(x-11\right)}=\dfrac{2x-19}{\left(x-9\right)\left(x-10\right)}\)

\(\Leftrightarrow\dfrac{2x-19}{x^2-19x+88}=\dfrac{2x-19}{x^2-19x+90}\)

do \(x^2-19x+88\ne x^2-19x+90\)

\(\Rightarrow2x-19=0\)

=> x=\(\dfrac{19}{2}\)

Vậy x=\(0\); x=\(\dfrac{19}{2}\)

Tik

X=0

Xin slot :)

\(x\ne8;9;10;11\)

\(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)

\(\Leftrightarrow\frac{x}{x-8}-\frac{x}{x-9}+\frac{x}{x-11}-\frac{x}{x-10}=0\)

\(\Leftrightarrow x\left(\frac{1}{x-8}-\frac{1}{x-9}+\frac{1}{x-11}-\frac{1}{x-10}\right)=0\)

\(\Leftrightarrow x\left(\frac{-1}{\left(x-8\right)\left(x-9\right)}+\frac{1}{\left(x-11\right)\left(x-10\right)}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{-1}{\left(x-8\right)\left(x-9\right)}+\frac{1}{\left(x-11\right)\left(x-10\right)}=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{1}{\left(x-8\right)\left(x-9\right)}=\frac{1}{\left(x-11\right)\left(x-10\right)}\Leftrightarrow\left(x-8\right)\left(x-9\right)=\left(x-11\right)\left(x-10\right)\)

\(\Leftrightarrow x^2-17x+72=x^2-21x+110\Leftrightarrow4x=38\Rightarrow x=\frac{19}{2}\)

\(\frac{x-7}{x-8}-\frac{x-8}{x-9}=\frac{x-10}{x-11}-\frac{x-11}{x-12}\)

\(\frac{x-7}{x-8}-\frac{x-8}{x-9}-\frac{x-10}{x-11}+\frac{x-11}{x-12}=0\)

Rồi còn lại làm típ