Question

cho số thực x thỏa mãn đk  \(0\le x\le1\)

tìm GTNN, GTLN của bt P=\(\frac{x^2}{2-x^2}+\frac{1-x^2}{1+x^2}\)

Answer 1

Đặt \(x^2=p\left(0\le p\le1\right)\)

Ta có : \(P=\frac{p}{2-p}+\frac{1-p}{1+p}=-2+\frac{2}{2-p}+\frac{2}{1+p}\)

\(=-2+2\left(\frac{1}{2-p}+\frac{1}{1+p}\right)=2\left(\frac{3}{\left(2-p\right)\left(1+p\right)}-1\right)\)

\(=2\left(\frac{3}{2+p\left(1-p\right)}-1\right)\)

Do \(0\le p\le1\Rightarrow p\left(1-p\right)\ge0\) \(\Rightarrow P\le2\left(\frac{3}{2}-1\right)=1\) có MAX là 1

Ta có : \(p\left(1-p\right)\le\frac{\left(p+1-p\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow P\ge2\left(\frac{3}{2+\frac{1}{4}}-1\right)=\frac{2}{3}\)Có MIN là \(\frac{2}{3}\)