Đáp án: $2020$
Giải thích các bước giải:
Đặt $\sqrt{y}=t,t\ge0$
$\to P=-3x^2-4xt+16x-2t^2+12t+1998$
$\to -P=3x^2+4xt-16x+2t^2-12t-1998$
$\to -P=3x^2+4x(t-4)+2t^2-12t-1998$
$\to -P=3(x^2+\dfrac43x(t-4))+2t^2-12t-1998$
$\to -P=3(x^2+2x\cdot \dfrac23(t-4)+(\dfrac23(t-4))^2)-3(\dfrac23(t-4))^2+2t^2-12t-1998$
$\to -P=3(x+\dfrac23(t-4))^2+\dfrac{2t^2-4t-64}{3}-1998$
$\to -P=3(x+\dfrac23(t-4))^2+\dfrac{2(t-1)^2-66}{3}-1998$
$\to -P=3(x+\dfrac23(t-4))^2+\dfrac23(t-1)^2-2020$
$\to -P\ge-2020$
$\to P\le 2020$
Dấu = xảy ra khi $x+\dfrac23(t-4)=t-1=0\to t=1, x=2\to y=1,x=2$
