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: \(x\ne\left\{3;4;5;6\right\}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\)và \(5x+y-2z=28\)

a, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)

=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1998}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)

=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}=0\right)\)

=> \(x-2004=0\)

=> \(x=2004\)

Vậy phương trình có nghiệm là x = 2004 .

b, Ta có : \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)

=> \(\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=10-1-2-3-4=0\)

=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)

=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)

=> \(x-100=0\)

=> \(x=100\)

Vậy phương trình có nghiệm là x = 100 .

x=-1               

x=-1                                 

x = -1

ai tk mk

mk tk lại

hứa luôn

thank nhiều

Công mỗi phân số cho 1 .....................

 mỗi hạng tử ở 2 vế cộng với 1 (có nghĩa là cộng 2 vế với 3 xong chia đều ra 3 hạng tử mỗi hạng tử cộng với 1)

Sau đó sẽ dẫn đến tất cả các hạng tử đều có chung tử số rồi nhóm tử ra ngoài là được 

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

Cộng mỗi p/s cho 1,ta đc:

\(\frac{x-6}{7}+1+\frac{x-7}{8}+1+\frac{x-8}{9}+1=\frac{x-9}{10}+1+\frac{x-10}{11}+1+\frac{x-11}{12}+1\)

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

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\left(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\right)=0\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\)

=>x+1=0

=>x=-1

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

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

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

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

\(\Rightarrow x\left(x-8+x-11-x+9-x+10\right)=0\)

\(\Rightarrow x.0=0\)

Vậy x thỏa mãn với mọi giá trị.

Câu còn lại bn lm tương tự nhé........ 

DKXD: x khác 3;4;5;6

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

\(\Leftrightarrow\frac{x^2-5x-x^2+3x}{\left(x-3\right).\left(x-5\right)}-\frac{x^2-6x-x^2+4x}{\left(x-4\right).\left(x-6\right)}=0\)

\(\Leftrightarrow\frac{2x}{\left(x-4\right).\left(x-6\right)}-\frac{2x}{\left(x-3\right).\left(x-5\right)}=0\)

\(\Leftrightarrow2x.\left(\frac{\left(x-3\right).\left(x-5\right)-\left(x-4\right).\left(x-6\right)}{\left(x-4\right).\left(x-6\right).\left(x-3\right).\left(x-5\right)}\right)=0\)

\(\Leftrightarrow2x.\left(\frac{2x-9}{\left(x-4\right).\left(x-5\right).\left(x-3\right).\left(x-6\right)}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{9}{2}\end{cases}}}\)

Vậy x=0 hoặc x=9/2

đéo biêt