$a,ĐKXĐ:a$$\geq0,a$ $\neq1$
$P=(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}).$ $\dfrac{1-\sqrt{a}}{1-a}$
$P=(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+$ $\dfrac{(1-\sqrt{a})\sqrt{a}}{1-\sqrt{a}}).$ $\dfrac{1-\sqrt{a}}{(1-\sqrt{a)}.(1+\sqrt{a)}}$
$P=(\dfrac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}).$ $\dfrac{1}{1+\sqrt{a}}$
$P=\dfrac{(1-a)+\sqrt{a}(1-a)}{1-\sqrt{a}}.$ $\dfrac{1}{1+\sqrt{a}}$
$P=\dfrac{(1-a)(1+\sqrt{a})}{1-\sqrt{a}}.$ $\dfrac{1}{1+\sqrt{a}}$
$P=\dfrac{1-a}{1-\sqrt{a}}$
$P=\dfrac{(1-\sqrt{a})(1+\sqrt{a})}{1-\sqrt{a}}$
$P={1+\sqrt{a}}$
$b,$
$P=2a⇔{1+\sqrt{a}}=2a$
$⇔1-2a+{\sqrt{a}}=0$
$⇔1-a-a+{\sqrt{a}}=0$$\geq0$
$⇔(1+{\sqrt{a}})(1-{\sqrt{a}})+{\sqrt{a}(1-{\sqrt{a}})=0}$
$⇔(1-{\sqrt{a})(1+{\sqrt{a}}+{\sqrt{a})}=0}$
$⇔(1-{\sqrt{a})(1+{2\sqrt{a})}=0}$
$⇔$\(\left[ \begin{array}{l}1-{\sqrt{a}=0}\\1+{2\sqrt{a}=0}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}a=1(L)\\a={∅}\end{array} \right.\)
Vậy $a={∅}$
