a: \(A=\frac{5+\sqrt5}{\sqrt5+1}-\sqrt{6-2\sqrt5}\)
\(=\frac{\sqrt5\left(\sqrt5+1\right)}{\sqrt5+1}-\sqrt{\left(\sqrt5-1\right)^2}\)
\(=\sqrt5-\left(\sqrt5-1\right)=1\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\frac{x-1}{x}\)
b: B<A
=>\(\frac{x-1}{x}-1<0\)
=>\(\frac{x-1-x}{x}<0\)
=>\(\frac{-1}{x}<0\)
=>x>0
Kết hợp ĐKXĐ, ta được: x>0 và x<>1
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