Đặt \(a=x^2;b=y^2\left(a;b\ge0\right)\)

\(A=\frac{\left(a-b\right)\left(1-ab\right)}{\left(1+a\right)^2\left(1+b\right)^2}\)

\(\left|A\right|=\frac{\left|\left(a-b\right)\left(1-ab\right)\right|}{\left(1+a\right)^2\left(1+b^2\right)}\le\frac{\left(a+b\right)\left(1+ab\right)}{\left(1+a\right)^2\left(1+b\right)^2}\)

\(\left(1+a\right)\left(1+b\right)=\left(a+b\right)+\left(1+ab\right)\ge2\sqrt{\left(a+b\right)\left(1+ab\right)}\)

\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\ge4\left(a+b\right)\left(1+ab\right)\)

\(\Rightarrow\left|A\right|\le4\)

\(\Rightarrow-4\le A\le4\)

\(A=-4\Leftrightarrow a=0;b=1\Leftrightarrow x=0;y=+1or-1\)

\(A=4\Leftrightarrow a=1;b=0\Leftrightarrow x=+-1;y=0\)

Vậy \(MinA=-4;MaxA=4\)

\(x^4+2x^2y^2-3x^2+y^4-4y^2+4=1\)

\(\Leftrightarrow\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)+4=1-x^2\)

\(\Leftrightarrow\left(x^2+y^2-2\right)^2=1-x^2\le1\)

\(\Rightarrow-1\le x^2+y^2-2\le1\)

\(\Rightarrow1\le x^2+y^2\le3\)

\(A_{min}=1\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)

\(A_{max}=0\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{3}\end{matrix}\right.\)

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pt <=> \(\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)+3=-x^2\le0\) (1)

(1)<=> \(A^2-4A+3\le0\Leftrightarrow1\le A\le3\)

Vậy GTNN của A là  1 tại x = 0 y =+- 1

GTLN của  A là 3 tại x = 0 ; y= +-căn3

thank nha thắng