\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4x}{x^2-1}\) (1)
điều kiện xác định: \(x\ne\pm1\)
(1) => \(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4x}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+1+x-1\right)\left(x+1-x+1\right)-4x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x.2-4x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{0x}{\left(x-1\right)\left(x+1\right)}=0\)
Vậy phương trình có nghiệm với mọi x \(\ne\pm1\)
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4x}{x^2-1}\)đkxđ \(x\ne\pm1\)
\(\Leftrightarrow x^2+2x+1-x^2-2x-1-4x=0\)
\(\Leftrightarrow-4x=0\)
\(\Leftrightarrow x=0\)
\(ĐKXĐ:x\ne\pm1\)
\(pt\Leftrightarrow\frac{\left(x+1\right)^2-\left(x-1\right)^2}{x^2-1}=\frac{4x}{x^2-1}\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4x\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4x\)
\(\Leftrightarrow4x=4x\)
Vậy pt đúng với mọi x khác 1 và -1
