ĐKXĐ: \(a\ge0;a\ne16\) rút gọn: A=\(\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{8+2\sqrt{a}-a}+\dfrac{\sqrt{a}+4}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{4-\sqrt{a}}=\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{8+4\sqrt{a}-2\sqrt{a}-a}+\dfrac{\sqrt{a}+4}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{4-\sqrt{a}}=\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{4\left(2+\sqrt{a}\right)-\sqrt{a}\left(2+\sqrt{a}\right)}+\dfrac{\sqrt{a}+4}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{4-\sqrt{a}}=\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\left(4-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{\sqrt{a}+4}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{4-\sqrt{a}}=\dfrac{-\sqrt{a}\left(2\sqrt{a}+1\right)+a-16+a+4\sqrt{a}+4}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+2\right)}=\dfrac{-2a-\sqrt{a}+2a+4\sqrt{a}-12}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+2\right)}=\dfrac{3\sqrt{a}-12}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+2\right)}=\dfrac{3\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+2\right)}=\dfrac{3}{\sqrt{a}+2}\)
Bạn ghi nhầm đề đúng ko, mk sửa lại đề đúng nè
