Đáp án: $6$
Giải thích các bước giải:
Đặt $x^2=a, y^2=b, (a , b\ge 0)$
$\to a+b=2$
$\to a=2-b$
Ta có:
$S=3x^2+5x^2y^2+2y^4+2y^2$
$\to S=3a+5ab+2b^2+2b$
$\to S=3(2-b)+5(2-b)b+2b^2+2b$
$\to S=6-3b+10b-5b^2+2b^2+2b$
$\to S=6+9b-3b^2$
$\to S=-3(b^2-3b)+6$
$\to S=-3(b^2-2b\cdot \dfrac32+\dfrac94)+6+\dfrac{27}{4}$
$\to S=-3(b-\dfrac32)^2+6+\dfrac{27}{4}$
Mà $a+b=2$ với $a, b\ge 0$
$\to 0\le b\le 2$
$\to -\dfrac32\le b-\dfrac32\le \dfrac12$
$\to 0\le (b-\dfrac32)^2\le (-\dfrac32)^2$
$\to 0\le (b-\dfrac32)^2\le \dfrac94$
$\to 0\le 3(b-\dfrac32)^2\le \dfrac{27}4$
$\to -\dfrac{27}{4}\le -3(b-\dfrac32)^2\le 0$
$\to 6\le -3(b-\dfrac32)^2+6+\dfrac{27}{4}\le 6+\dfrac{27}{4}$
$\to 6\le S\le 6+\dfrac{27}{4}$
$\to GTNN_S=6$ khi đó $b=2\to a=0$
$\to x^2=0, y^2=2$
