\(P=\frac{x^2}{2-x^2}+\frac{1-x^2}{1+x^2}\)
\(\Leftrightarrow P=\frac{x^2-2+2}{2-x^2}+\frac{-1-x^2+2}{1+x^2}\)
\(\Leftrightarrow P=-1+\frac{2}{2-x^2}-1+\frac{2}{1+x^2}\)
\(\Leftrightarrow P=-1-1+2\left(\frac{1}{2-x^2}+\frac{1}{1+x^2}\right)\)
Ta sẽ c/m \(\frac{1}{2-x^2}+\frac{2}{1+x^2}\le\frac{3}{2}\)
\(\frac{1}{2-x^2}+\frac{1}{1+x^2}\le\frac{3}{2}\)
\(\Leftrightarrow\frac{1+x^2+2-x^2}{\left(2-x^2\right)\left(1+x^2\right)}\le\frac{3}{2}\)
\(\Leftrightarrow\frac{3}{\left(2-x^2\right)\left(1+x^2\right)}\le\frac{3}{2}\)
\(\Leftrightarrow\frac{1}{\left(2-x^2\right)\left(1+x^2\right)}\le\frac{1}{2}\)
\(\Leftrightarrow2\le2+2x^2-x^2-x^4\)
\(\Leftrightarrow0\le x^2-x^4\)
\(\Leftrightarrow0\le x^2\left(1-x^2\right)\)( luôn đúng với \(0\le x\le1\))
Dấu " = " xảy ra \(\Leftrightarrow\orbr{\begin{cases}x^2=0\\1-x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
=> \(P\le-1-1+2.\frac{3}{2}=-2+3=1\)
Dấu " = " xảy ra \(\Leftrightarrow\orbr{\begin{cases}x^2=0\\1-x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(P_{max}=1\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
P/S: có gì sai sót xin bỏ qua
