a: ĐKXĐ: a>0; a<>1
\(A=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
\(=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}-1+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+1}=\frac{a-1}{\sqrt{a}}\)
b: Thay \(a=4+2\sqrt3=\left(\sqrt3+1\right)^2\) vào A, ta được:
\(A=\frac{4+2\sqrt3-1}{\sqrt{\left(\sqrt3+1\right)^2}}=\frac{3+2\sqrt3}{\sqrt3+1}=\frac{\left(2\sqrt3+3\right)\left(\sqrt3-1\right)}{\left(\sqrt3+1\right)\left(\sqrt3-1\right)}\)
\(=\frac{6-2\sqrt3+3\sqrt3-3}{3-1}=\frac{3+\sqrt3}{2}\)
c: A<0
=>\(\frac{a-1}{\sqrt{a}}\) <0
=>a-1<0
=>a<1
=>0<a<1
