a) Ta có: \(A=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{3x+8\sqrt{x}-3\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}\left(3\sqrt{x}+8\right)-\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(3\sqrt{x}+8\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b) Ta có: \(x=3+2\sqrt{2}\)
\(=1+2\cdot1\cdot\sqrt{2}+2\)
\(=\left(1+\sqrt{2}\right)^2\)
Thay \(x=\left(1+\sqrt{2}\right)^2\) vào biểu thức \(A=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\), ta được:
\(A=\frac{3\cdot\sqrt{\left(1+\sqrt{2}\right)^2}+8}{\sqrt{\left(1+\sqrt{2}\right)^2}+2}\)
\(=\frac{3\cdot\left|1+\sqrt{2}\right|+8}{\left|1+\sqrt{2}\right|+2}\)
\(=\frac{3\left(1+\sqrt{2}\right)+8}{1+\sqrt{2}+2}\)
\(=\frac{3+3\sqrt{2}+8}{3+\sqrt{2}}\)
\(=\frac{11+3\sqrt{2}}{3+\sqrt{2}}=\frac{27-2\sqrt{2}}{7}\)
Vậy: Khi \(x=3+2\sqrt{2}\) thì \(A=\frac{27-2\sqrt{2}}{7}\)
