Đáp án:
ĐKXĐ: $y≥0,\,y\ne1$
Rút gọn: $\dfrac{2}{1-y}$
Giải thích các bước giải:
$\Bigg(\dfrac{1}{1-\sqrt{y}}+\dfrac{1}{1+\sqrt{y}}\Bigg):\dfrac{1}{1-\sqrt{y}}-\dfrac{1}{1+\sqrt{y}}+\dfrac{1}{1-\sqrt{y}}$
ĐKXĐ: $\begin{cases}y≥0\\1-\sqrt{y}\ne0\end {cases} \to y≥0,\,y\ne1$
$=\dfrac{1+\sqrt{y}+1-\sqrt{y}}{(1-\sqrt{y})(1+\sqrt{y})}.(1-\sqrt{y})-\dfrac{1-\sqrt{y}-1-\sqrt{y}}{(1-\sqrt{y})(1+\sqrt{y})}$
$=\dfrac{2}{1+\sqrt{y}}-\dfrac{-2\sqrt{y}}{(1-\sqrt{y})(1+\sqrt{y})}$
$=\dfrac{2(1-\sqrt{y})+2\sqrt{y}}{(1-\sqrt{y})(1+\sqrt{y})}$
$=\dfrac{2}{(1-\sqrt{y})(1+\sqrt{y})}$
$=\dfrac{2}{1-y}$
