Thực hiện phép tính: $\frac{x+3}{x+1}$-$\frac{2x-1}{x-1}$- $\frac{x}{x^{2}-1 }$

Đáp án+Giải thích các bước giải:

$\begin{array}{l}\dfrac{x+3}{x+1}-\dfrac{2x-1}{x-1}-\dfrac{x}{x^2-1}\\ĐKXĐ:x \neq 1,-1\\=\dfrac{(x+3)(x-1)}{(x-1)(x+1)}-\dfrac{(2x-1)(x+1)}{(x-1)(x+1)}-\dfrac{x}{(x-1)(x+1)}\\=\dfrac{x^2+2x-3}{(x-1)(x+1)}-\dfrac{2x^2+x-1}{(x-1)(x+1)}-\dfrac{x}{(x-1)(x+1)}\\=\dfrac{x^2+2x-3-2x^2-x+1-x}{(x-1)(x+1)}\\=\dfrac{-x^2+1}{(x-1)(x+1)}\\=\dfrac{-(x^2-1)}{(x-1)(x+1)}\\=\dfrac{-(x+1)(x-1)}{(x-1)(x+1)}\\=-1\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$