\(a,Đkxđ:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(P=\left(\frac{2a+1}{\sqrt{a^3}-1}-\frac{\sqrt{a}}{a+\sqrt{a}+1}\right)\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\frac{2a+1-\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\left(a-\sqrt{a}+1-\sqrt{a}\right)\)
\(=\frac{2a+1-a+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\left(\sqrt{a}-1\right)^2\)
\(=\frac{a+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\left(\sqrt{a}-1\right)^2\)
\(=\sqrt{a}-1\)
\(b,\) Ta có: \(P\sqrt{1-a}=\left(\sqrt{a}-1\right)\sqrt{1-a}\) với \(a\ge0\) và \(a< 1\)
\(\Rightarrow\sqrt{a}< 1\)
\(\Rightarrow\sqrt{a}-1< 0\)
\(\Rightarrow P\sqrt{1-a}< 0\)
Vậy ............
