điều kiện xác định : \(a\ge0;a\ne1\)
a) ta có : \(P=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow P=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow P=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
b) ta có : \(P>-\sqrt{2}\Leftrightarrow1-a>-\sqrt{2}\Leftrightarrow a< \sqrt{2}+1\)
c) để \(P=\sqrt{a}\Leftrightarrow1-a=\sqrt{a}\Leftrightarrow-a-\sqrt{a}+1=0\)
giải đen ta ta có : \(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{-1+\sqrt{5}}{2}\\a=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\) (tmđk)
