Question

Giải phương trình sau :

\(\frac{1}{x+1}-\frac{1}{x-1}=\frac{3x-6}{1-x^2}\)

Answer 1

\(\frac{1}{x+1}+\frac{1}{1-x}=\frac{3x-6}{1-x^2}\)

\(\frac{1-x+x+1}{1-x^2}=\frac{3x-6}{1-x^2}\)

\(2=3x-6\)

\(4=3x\)

\(x=\frac{4}{3}\)

Answer 2

\(\frac{1}{x+1}-\frac{1}{x-1}=\frac{3x-6}{1-x^2}\)

\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{1-x}=\frac{3x-6}{\left(1-x\right)\left(x+1\right)}\)

Quy đồng rồi khử mẫu ta được:

\(1-x+x+1=3x-6\)

\(\Leftrightarrow-x+x-3x=-6-1-1\)

\(\Leftrightarrow-3x=-8\)

\(\Leftrightarrow x=\frac{8}{3}\)

Vậy  .... 

Answer 3

Cảm ơn nhé

Answer 4

sửa cho mình cái kq là 8/3 nha bn iu, mình tính nhầm

Answer 5

ok bn linh chi

Answer 6

\(\frac{1}{x+1}-\frac{1}{x-1}=\frac{3x-6}{1-x^2}\)

\(\Leftrightarrow\frac{x-1}{x^2-1}-\frac{x+1}{x^2-1}=-\frac{3x-6}{x^2-1}\)

\(\Rightarrow x-1-x-1=-3x+6\)

\(\Leftrightarrow-2+3x-6=0\)

\(\Leftrightarrow3x-8=0\Leftrightarrow x=\frac{8}{3}\)

Answer 7

\(\frac{1}{x+1}-\frac{1}{x-1}=\frac{3x-6}{1-x^2}\)  \(\left(ĐKXĐ:x\ne\pm1\right)\)

\(\Leftrightarrow\frac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x-6}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Rightarrow\left(x-1\right)-\left(x+1\right)+\left(3x-6\right)=0\)

\(\Leftrightarrow x-1-x-1+3x-6=0\)

\(\Leftrightarrow3x-8=0\)

\(\Leftrightarrow x=\frac{8}{3}\left(tm\right)\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{8}{3}\right\}\)