Đáp án: $A=\dfrac{4y}{3-\sqrt{y}}$
Giải thích các bước giải:
Ta có:
$A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)$
$\to A=\left(\dfrac{4\sqrt{y}\left(2-\sqrt{y}\right)}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}+\dfrac{8y}{\left(2-\sqrt{y}\right)\left(2+\sqrt{y}\right)}\right):\left(\dfrac{\sqrt{y}-1}{\sqrt{y}\left(\sqrt{y}-2\right)}-\dfrac{2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)$
$\to A=\left(\dfrac{8\sqrt{y}-4y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}+\dfrac{8y}{\left(2-\sqrt{y}\right)\left(2+\sqrt{y}\right)}\right):\left(\dfrac{\sqrt{y}-1}{\sqrt{y}\left(\sqrt{y}-2\right)}-\dfrac{2\sqrt{y}-4}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)$
$\to A=\left(\dfrac{8\sqrt{y}-4y+8y}{\left(2-\sqrt{y}\right)\left(2+\sqrt{y}\right)}\right):\left(\dfrac{\sqrt{y}-1-2\sqrt{y}+4}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)$
$\to A=\left(\dfrac{8\sqrt{y}+4y}{\left(2-\sqrt{y}\right)\left(2+\sqrt{y}\right)}\right):\left(\dfrac{3-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-2\right)}\right)$
$\to A=\dfrac{4\sqrt{y}\left(2+\sqrt{y}\right)}{\left(2-\sqrt{y}\right)\left(2+\sqrt{y}\right)}\cdot\dfrac{\sqrt{y}\left(\sqrt{y}-2\right)}{3-\sqrt{y}}$
$\to A=\dfrac{4y}{3-\sqrt{y}}$
