Question

\(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right].\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a< >1\end{matrix}\right.\)

\(\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right)\cdot\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)

\(=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{a\sqrt{a}}{a-1}\right)\cdot\dfrac{\sqrt{a}-1+\sqrt{a}+1}{a-1}\)

\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{a-1}\cdot\dfrac{2\sqrt{a}}{a-1}\)

\(=\dfrac{2\sqrt{a}\left(a+2\sqrt{a}+1-a\sqrt{a}\right)}{\left(a-1\right)^2}\)