\(\left(x^2+y^2+1\right)^2+1\le\left(x^2+y^2+1\right)^2+3x^2y^2+1\le4x^2+5y^2\le5x^2+5y^2\)
\(\Rightarrow\left(x^2+y^2+1\right)^2-5\left(x^2+y^2+1\right)+6\le0\)
\(\Leftrightarrow2\le x^2+y^2+1\le3\)
\(P=\frac{x^2+2y^2-3x^2y^2}{x^2+y^2+1}+3-3=\frac{\left(x^2+y^2+1\right)^2+4}{x^2+y^2+1}-3\)
Đặt \(x^2+y^2+1=t\Rightarrow t\in\left[2;3\right]\)
\(\Rightarrow P=\frac{t^2+4}{t}-3=t+\frac{4}{t}-3\ge2\sqrt{\frac{4t}{t}}-3=1\)
\(P_{min}=1\) khi \(\left(x^2;y^2\right)=\left(0;1\right)\)
\(P=\frac{t^2+4}{t}-\frac{13}{3}+\frac{4}{3}=\frac{3t^2-13t+12}{3t}+\frac{4}{3}=\frac{\left(t-3\right)\left(3t-4\right)}{3t}+\frac{4}{3}\le\frac{4}{3}\)
\(P_{max}=\frac{4}{3}\) khi \(\left(x^2;y^2\right)=\left(0;2\right)\)
tại sao P = t^2+4 vậy ạ? t^2 ra mũ 4 rồi mà
