giải giúp mk với mk cảm ơn

`ĐKXĐ: x ne +- 1`

$\dfrac{x^2+3x}{x^2-1}-\dfrac{x+1}{x-1}\\
=\dfrac{x^2+3x}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2+3x}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2+3x-x^2-2x-1}{\left(x-1\right)\left(x+1\right)}\\
=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\\
=\dfrac{1}{x+1}$