ĐK:a≥0 ; a≠1
a) $A=(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}):(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1})$
$=(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}(\sqrt{a}-1)}):(\frac{1}{\sqrt{a}+1}+\frac{2}{(\sqrt{a}-1)(\sqrt{a}+1)})$
$=\frac{\sqrt{a}.\sqrt{a}-1}{\sqrt{a}(\sqrt{a}-1)}:\frac{\sqrt{a}-1+2}{(\sqrt{a}+1)(\sqrt{a}-1)}$
$=\frac{a-1}{\sqrt{a}(\sqrt{a}-1)}.\frac{(\sqrt{a}+1)(\sqrt{a}-1)}{\sqrt{a}+1}$
$=\frac{a-1}{\sqrt{a}}$
b) Ta có : $a=4+2\sqrt{3}$
$=3+2\sqrt{3}+1$
$=(\sqrt{3}+1)^2$
Thay a bào biều thức A ta được : $\frac{4+2\sqrt{3}-1}{\sqrt{(\sqrt{3}+1)^2}}$
$=\frac{3+2\sqrt{3}}{\sqrt{3}+1}=\frac{(3+2\sqrt{3})(\sqrt{3}-1)}{3-1}$
$=\frac{3\sqrt{3}-3+6-2\sqrt{3}}{2}=\frac{3-\sqrt{3}}{2}$
c) Để A<0 thì $\frac{a-1}{\sqrt{a}}<0$
⇔$a-1<0$
⇔$a<1$ (loại)
Vậy A<0 thì a=∅
