A=(\(\dfrac{a+3\sqrt{a}+2}{\sqrt{a}+2\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\)):(\(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\))
B=\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}\dfrac{4}{\sqrt{a}}\right)\)
a, Tìm ĐKXĐ của \(\sqrt{a}\)
b, Rút gọn các biểu thức trên
a: ĐKXĐ: a>=0; a<>1
b: \(A=\left(\dfrac{a+3\sqrt{a}+2}{3\sqrt{a}-2}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}-1+\sqrt{a}+1}{a-1}\)
\(=\left(\dfrac{\left(a-1\right)\left(\sqrt{a}+2\right)-3a+2\sqrt{a}}{\left(3\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{a-1}{2\sqrt{a}}\)
\(=\dfrac{a\sqrt{a}+2a-\sqrt{a}-2-3a+2\sqrt{a}}{\left(3\sqrt{a}-2\right)}\cdot\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
\(=\dfrac{\left(a\sqrt{a}-a+\sqrt{a}-2\right)}{3\sqrt{a}-2}\cdot\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
\(B=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
