\(DK:\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
\(P=\left(\frac{a-1}{2\sqrt{a}}\right)\left[\frac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{-4a}{a-1}\right)=\frac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)
b) \(P=-a\Leftrightarrow-2\sqrt{a}=-a\)
\(\Leftrightarrow-2\sqrt{a}+\sqrt{a}.\sqrt{a}=0\)
\(\Leftrightarrow\sqrt{a}\left(-2+\sqrt{a}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=0\\\sqrt{a}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=4\end{matrix}\right.\)
Vậy \(P=-a\Leftrightarrow a=0;a=4\)
c) \(P>-2\Leftrightarrow-2\sqrt{a}>-2\)
\(\Leftrightarrow\sqrt{a}< 1\Leftrightarrow a< 1\)
Mà \(a\ge0\)
\(\Rightarrow a=0\)
Vậy \(P>-2\Leftrightarrow a=0\)
mn giúp mk nhanh vs nha vì tối mk đi hc r tks trước :))
