Toán [toán 9] thi học sinh giỏi

L

leminhnghia1

Giải:

ĐK: $x \not =0$

Ta có: $GT \iff x^4+8+\dfrac{x^2y^2}{8}=8x^2$

$\iff (x^4-8x^2+16)+\dfrac{(xy)^2}{8}=8$

$\iff (x^2-4)^2+\dfrac{(xy)^2}{8}=8$

$\rightarrow \dfrac{(xy)^2}{8} \le 8$

$\rightarrow -8 \le xy \le 8$

$\rightarrow Min=-8$

$P=xy+2024 \ge -8+2024=2016$

$Min=2016 \iff xy=-8; x^2=4$

$\rightarrow x=-2;y=4$ v $x=2;y=-4$
 
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