a: Ta có: \(A=\frac{8+2\sqrt2}{3-\sqrt2}-\frac{2+3\sqrt2}{\sqrt2}+\frac{\sqrt2}{1-\sqrt2}\)
\(=\frac{\left(8+2\sqrt2\right)\left(3+\sqrt2\right)}{\left(3-\sqrt2\right)\left(3+\sqrt2\right)}-3-\sqrt2-\sqrt2\left(\sqrt2+1\right)\)
\(=\frac{24+8\sqrt2+6\sqrt2+4}{7}-3-\sqrt2-2-\sqrt2\)
\(=\frac{28+14\sqrt2}{7}-5-2\sqrt2=4+2\sqrt2-5-2\sqrt2=-1\)
b: \(B=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+4\sqrt{x}\right)\cdot\frac{x-1}{\sqrt{x}}\)
\(=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\frac{x-1}{\sqrt{x}}=4\sqrt{x}\left(\frac{1}{x-1}+1\right)\cdot\frac{x-1}{\sqrt{x}}\)
\(=4\left(x-1\right)\cdot\frac{1+x-1}{x-1}=4\left(x-1\right)\cdot\frac{x}{x-1}=4x\)
