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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 \(\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{3}\end{matrix}\right.\)
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
\(x^4+2x^2y^2+y^4-3x^2-4y^2+4=1\)
\(\Leftrightarrow\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)+4=1-x^2\)
\(\Leftrightarrow\left(x^2+y^2-2\right)^2=1-x^2\)
Do \(1-x^2\le1\) \(\forall x\)
\(\Rightarrow-1\le x^2+y^2-2\le1\)
\(\Rightarrow1\le x^2+y^2\le3\)
\(A_{min}=1\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)
\(A_{max}=3\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{3}\end{matrix}\right.\)
\(A\le\left|x\right|+\sqrt{2}+\left|y\right|+1=6+\sqrt{2}\)
\(A_{max}=6+\sqrt{2}\) khi \(\left\{{}\begin{matrix}x\le0\\y\le0\\\left|x\right|+\left|y\right|=5\end{matrix}\right.\)
\(A\ge\left|x+y-\sqrt{2}-1\right|\ge4-\sqrt{2}\)
\(A_{min}=4-\sqrt{2}\) khi \(\left\{{}\begin{matrix}x\ge\sqrt{2}\\y\ge1\\x+y=5\end{matrix}\right.\)
2/ \(A\ge\frac{1}{3}\left(x^2+y^2+z^2\right)^2\ge\frac{1}{3}\left(xy+yz+zx\right)^2=\frac{1}{3}\)
\(A_{min}=\frac{1}{3}\) khi \(x=y=z=\frac{1}{\sqrt{3}}\)
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