tìm min max

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vansang02121998

$x^2(x^2+2y^2-3)+(y^2-2)^2=1$

$\Leftrightarrow (x^2+y^2-2)^2=1-x^2 \le 1$

$\Leftrightarrow -1 \le x^2+y^2-2 \le 1$

$\Leftrightarrow 1 \le x^2+y^2 \le 3$

$min(x^2+y^2)=1 \Leftrightarrow x=0; |y|=1$

$max(x^2+y^2)=3 \Leftrightarrow x=0; |y|=\sqrt{3}$




Lười gõ: $3-2\sqrt{2} \le \dfrac{x^2-2x+2}{x^2+2x+2} \le 3+2\sqrt{2}$
 
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