\(\left(x^2-y^2+1\right)^2+4x^2y^2-x^2-y^2=0\)
\(\Leftrightarrow\left[\left(x^2+1\right)-y^2\right]^2+4x^2y^2-x^2-y^2=0\)
\(\Leftrightarrow x^4+2x^2+1+y^4-2y^2\left(x^2+1\right)+4x^2y^2-x^2-y^2=0\)
\(\Leftrightarrow x^4+y^4+2x^2y^2+x^2-3y^2+1=0\)
\(\Leftrightarrow x^4+y^4+2x^2y^2-3x^2-3y^2+1=-4x^2\)
\(\Leftrightarrow\left(x^2+y^2\right)^2-3\left(x^2+y^2\right)+1=-4x^2\)
Đặt t = x2 + y2 . Ta có : t2 - 3t +1 = -4x2
Suy ra : \(t^2-3t+1\le0\)
\(\Leftrightarrow t^2-2.\dfrac{3}{2}.t+\dfrac{9}{4}-\dfrac{5}{4}\le0\)
\(\Leftrightarrow\left(t-\dfrac{3}{2}\right)^2\le\dfrac{5}{4}\Leftrightarrow\left|t-\dfrac{3}{2}\right|\le\dfrac{\sqrt{5}}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{5}}{2}\le t-\dfrac{3}{2}\le\dfrac{\sqrt{5}}{2}\)
\(\Leftrightarrow\dfrac{3-\sqrt{5}}{2}\le t\le\dfrac{3+\sqrt{5}}{2}\)
Vì t = x2 + y2 nên :
GTNN của x2 + y2 là :\(\dfrac{3-\sqrt{5}}{2}\)
GTLN của x2 + y2 là :\(\dfrac{3+\sqrt{5}}{2}\)
