a/ đk\(-1\le a\le1\)
P\(=\left[\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right]:\left(\frac{3}{\sqrt{1-a^2}}+1\right)\)
\(=\left[\frac{3}{\sqrt{1+a}}+\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\right]:\) \(\left(\frac{3}{\sqrt{1-a^2}}+\frac{\sqrt{1-a^2}}{\sqrt{1-a^2}}\right)\)
=\(\frac{3+\sqrt{1-a^2}}{\sqrt{1+a}}.\frac{\sqrt{1+a}.\sqrt{1-a}}{3+\sqrt{1-a^2}}\)\(=\sqrt{1-a}\)
vậy \(P=\sqrt{1-a}\) với\(-1\le a\le1\)
b/ để P=\(\frac{1}{\sqrt{1-a^2}}\) thì: \(\sqrt{1-a}=\frac{1}{\sqrt{1-a}.\sqrt{1+a}}\)
\(\Leftrightarrow1-a=\frac{1}{1-a^2}\)
\(\Leftrightarrow a^3-a^2-a+1=1\) \(\Leftrightarrow a\left(a^2-a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\left(TM\right)\\a^2-a-1=0\end{matrix}\right.\) \(\Leftrightarrow\left(a-\frac{1}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}a-\frac{1}{2}=\frac{\sqrt{5}}{2}\\a-\frac{1}{2}=\frac{-\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{\sqrt{5}+1}{2}\left(loại\right)\\a=\frac{1-\sqrt{5}}{2}\left(TM\right)\end{matrix}\right.\)
vậy \(a=\frac{1-\sqrt{5}}{2}\) thì P=\(\frac{1}{\sqrt{1-a^2}}\)