a) Ta có: \(P=\frac{a+4\sqrt{a}+4}{\sqrt{a}+2}-\frac{4-a}{\sqrt{a}-2}\)
\(=\left(\sqrt{a}+2\right)+\frac{a-4}{\sqrt{a}-2}\)
\(=\sqrt{a}+2+\sqrt{a}+2\)
\(=2\sqrt{a}+4\)
b) Ta có: \(a=\frac{1}{3-2\sqrt{2}}+\frac{1}{3+2\sqrt{2}}\)
\(=\frac{1}{1-2\sqrt{2}+2}+\frac{1}{1+2\sqrt{2}+2}\)
\(=\frac{1}{\left(1-\sqrt{2}\right)^2}+\frac{1}{\left(1+\sqrt{2}\right)^2}\)
\(=\frac{\left(1+\sqrt{2}\right)^2}{\left(1-\sqrt{2}\right)^2\cdot\left(1+\sqrt{2}\right)^2}+\frac{\left(1-\sqrt{2}\right)^2}{\left(1-\sqrt{2}\right)^2\cdot\left(1+\sqrt{2}\right)^2}\)
\(=\frac{1+2\sqrt{2}+2+1-2\sqrt{2}+2}{\left(1-\sqrt{2}\right)^2\cdot\left(1+\sqrt{2}\right)^2}\)
\(=6\)
Thay a=6 vào biểu thức \(P=2\sqrt{a}+4\), ta được:
\(P=2\cdot\sqrt{6}+4\)
Vậy: Khi \(a=\frac{1}{3-2\sqrt{2}}+\frac{1}{3+2\sqrt{2}}\) thì \(P=2\cdot\sqrt{6}+4\)