\(\left(\frac{1}{x-2}-\frac{1}{x+2}\right)+\left(\frac{1}{x-1}-\frac{1}{x+1}\right)=0\)
\(\frac{x+2-x+2}{x^2-4}+\frac{x+1-x+1}{x^2-1}=0\)
\(\frac{4}{x^2-4}+\frac{2}{x^2-1}=0\)
\(4x^2-4+2x^2-8=0\)
\(6x^2-12=0\)
\(x^2=2\)
\(x=\sqrt{2}\)
ĐKXĐ: x
≠-2,-1,1,2
Ta có :
\(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+1}+\frac{1}{x+2}\)
<=> \(\frac{1}{x-1}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x-2}\)
<=>\(\frac{2}{x^2-1}=\frac{-4}{x^2-4}\)
<=> \(2x^2-8=-4x^2+4\)
<=> \(6x^2=12\)
<=> \(x^2=2\)
<=>\(\hept{\begin{cases}x=\sqrt{2}\left(TMĐK\right)\\x=-\sqrt{2}\left(TMĐK\right)\end{cases}}\)
Vậy pt trên có tập nghiệm S={\(\sqrt{2},-\sqrt{2}\)}
k mk nha mn
