Đáp án:
Giải thích các bước giải:
$\dfrac{x(√x+1)}{x-1}$ - $\dfrac{2x-√x}{x-√x}$
= $\dfrac{x(√x+1)}{(√x-1)(√x+1)}$ - $\dfrac{2x-√x}{√x(√x-1)}$
= $\dfrac{x.√x(√x+1)}{√x(√x-1)(√x+1)}$ - $\dfrac{(2x-√x)(√x+1)}{√x(√x-1)}$
= $\dfrac{x√x(√x+1) - (2x-√x)(√x+1)}{√x(√x-1)(√x+1)}$
= $\dfrac{(√x+1)(x√x-2x+√x)}{√x(√x-1)(√x+1)}$
= $\dfrac{(x√x-2x+√x)}{√x(√x-1)}$
