a)ĐKXĐ : \(a,b\ge0,a\ne b\)
Ta có \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a+1\right)\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{ab}+b}=\left(\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+a+\sqrt{ab}+b}{a\sqrt{a}-b\sqrt{b}}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}\right).\dfrac{a+\sqrt{ab}+b}{\left(a+1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{\left(a+1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{1}{a+1}\)
b)Ta có \(Q=P\left(3a+5\right)=\dfrac{3a+5}{a+1}=3+\dfrac{2}{a+1}\) nhận GT nguyên <=> \(\dfrac{2}{a+1}\) nhận GT nguyên <=> \(a+1\in\left\{-2,-1,1,2\right\}\) <=> \(a\in\left\{-3,-2,0,1\right\}\)
