a) \(A=\left(\frac{\sqrt{x}}{x\sqrt{x}-1}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(A=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(A=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) \(x=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(x=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(x=\left|\sqrt{2}+1\right|-\left|\sqrt{2}-1\right|\)
\(x=\sqrt{2}+1-\sqrt{2}+1=2\)
Thay vào \(A=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)
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