Điều kiện : \(a>0;a\ne1\).
a) Với mọi \(a>0;a\ne1\), ta có :
\(P=\left(\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}-2}{a-1}\right)\cdot\left(2+\dfrac{a+1}{\sqrt{a}}\right)\)
\(=\left[\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\cdot\dfrac{2\sqrt{a}+a+1}{\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
\(=\dfrac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}=\dfrac{2}{\sqrt{a}-1}\).
b) Để \(P< -1\) thì \(\dfrac{2}{\sqrt{a}-1}< -1\)
\(\Leftrightarrow\dfrac{2}{\sqrt{a}-1}+1< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-1}< 0\Rightarrow\sqrt{a}-1< 0\Leftrightarrow\sqrt{a}< 1\)
Suy ra : \(a< 1\). Kết hợp với điều kiện : Vậy \(P< -1\Leftrightarrow0< a< 1\).
