a: \(A=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)
\(=2+\dfrac{2a+2}{\sqrt{a}}=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
b: Để A=7 thì \(2a-5\sqrt{a}+2=0\)
\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(2\sqrt{a}-1\right)=0\)
=>a=4 hoặc a=1/4