Question

\(A=\left(\dfrac{\sqrt{a}}{2-\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}+2}-\dfrac{4a+2\sqrt{a}-4}{a-4}\right):\left(\dfrac{-2}{2-\sqrt{a}}+\dfrac{2+\sqrt{a}}{2\sqrt{a}-a}\right)\)

với a>0, a\(\ne4\)

a) rút gọn biểu thức A

b) tìm a để A= \(\sqrt{a}+2\)

Answer 1

a: \(=\dfrac{-a-2\sqrt{a}+a-2\sqrt{a}-4a-2\sqrt{a}+4}{a-4}:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\)

\(=\dfrac{-4a-6\sqrt{a}+4}{a-4}\cdot\dfrac{-\sqrt{a}\left(\sqrt{a}-2\right)}{-\sqrt{a}+2}\)

\(=\dfrac{4a+6\sqrt{a}-4}{\sqrt{a}+2}\cdot\dfrac{\sqrt{a}}{2-\sqrt{a}}=\dfrac{\sqrt{a}\left(4a+6\sqrt{a}-4\right)}{4-a}\)

b: Để \(A=\sqrt{a}+2\) thì \(4a\sqrt{a}+6a-4\sqrt{a}=\left(\sqrt{a}+2\right)\left(4-a\right)=4\sqrt{a}-a\sqrt{a}+8-2a\)

=>\(5a\sqrt{a}+8a-8\sqrt{a}-8=0\)

=>\(5a\cdot\sqrt{a}+10a-2a-4\sqrt{a}-4\sqrt{a}-8=0\)

=>\(\left(\sqrt{a}+2\right)\left(5a-2\sqrt{a}-4\right)=0\)

=>\(5a-2\sqrt{a}-4=0\)

=>\(a=\dfrac{22+2\sqrt{21}}{25}\)