a) Ta có: \(A=\left(\dfrac{\sqrt{x}+10}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{1+\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+10+x-4-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)
\(=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}\)
b) Thay \(x=3+2\sqrt{2}\) vào A, ta được:
\(A=\dfrac{\sqrt{2}+1+7}{\sqrt{2}+1+2}=\dfrac{8+\sqrt{2}}{3+\sqrt{2}}=\dfrac{22-5\sqrt{2}}{7}\)
c) Để \(A=\dfrac{2}{3}\) thì \(3\left(\sqrt{x}+7\right)=2\left(\sqrt{x}+2\right)\)
\(\Leftrightarrow3\sqrt{x}+21=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=-17\)(vô lý)
d) Để A>3 thì A-3>0
\(\Leftrightarrow\dfrac{\sqrt{x}+7}{\sqrt{x}+2}-3>0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+7-3\sqrt{x}-6}{\sqrt{x}+2}>0\)
\(\Leftrightarrow-2\sqrt{x}>-1\)
\(\Leftrightarrow\sqrt{x}< \dfrac{1}{2}\)
hay \(x< \dfrac{1}{4}\)
Kết hợp ĐKXĐ, ta được: \(0\le x< \dfrac{1}{4}\)
f) Để A nguyên thì \(\sqrt{x}+7⋮\sqrt{x}+2\)
\(\Leftrightarrow5⋮\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}+2\in\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow\sqrt{x}=3\)
hay x=9(thỏa ĐK)
