a) điều kiện \(a\ge0;a\ne1\)
\(P=\dfrac{\sqrt{a}-2}{1-\sqrt{a}}-\dfrac{1+\sqrt{a}}{2+\sqrt{a}}+\dfrac{3a-3+\sqrt{9a}}{a+\sqrt{a}-2}\)
\(P=\dfrac{2-\sqrt{a}}{\sqrt{a}-1}-\dfrac{1+\sqrt{a}}{2+\sqrt{a}}+\dfrac{3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\)
\(P=\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)-\left(1+\sqrt{a}\right)\left(\sqrt{a}-1\right)+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\)
\(P=\dfrac{4-a-\left(a-1\right)+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\)
\(P=\dfrac{4-a-a+1+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\) \(=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(2+\sqrt{a}\right)}\)
\(P=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)
b) điều kiện \(a\in Z;x\ge0;x\ne1\)
ta có : \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\) nguyên
\(\Leftrightarrow\sqrt{x}-1\) thuộc ước của 2 là : \(\pm1;\pm2\)
ta có : * \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tmđk\right)\)
* \(\sqrt{x}-1=-1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(tmđk\right)\)
* \(\sqrt{x}-1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(loại\right)\)
\(\sqrt{x}-1=-2\Leftrightarrow\sqrt{x}=-1\left(vôlí\right)\)
vậy \(x=4;x=0\)
