ĐKXĐ: \(x\ne ±1\)
Đặt \(A=\dfrac{1-\dfrac{2}{x+1}}{1-\dfrac{x^2-2}{x^2-1}}\)
\(1-\dfrac{2}{x+1}\\=\dfrac{x+1-2}{x+1}\\=\dfrac{x-1}{x+1}\)
\(1-\dfrac{x^2-2}{x^2-1}\\=\dfrac{x^2-1-x^2+2}{x^2-1}\\=\dfrac{1}{x^2-1}\\=\dfrac{1}{(x-1)(x+1)}\)
\(→A=\dfrac{\dfrac{x-1}{x+1}}{\dfrac{1}{(x-1)(x+1)}}\\=\dfrac{x-1}{x+1}:\dfrac{1}{(x-1)(x+1)}\\=\dfrac{x-1}{x+1}.(x-1)(x+1)\\=(x-1)^2\)
Vậy với \(x\ne ±1\) thì \(\dfrac{1-\dfrac{2}{x+1}}{1-\dfrac{x^2-2}{x^2-1}}=(x-1)^2\)
