Đáp án: $b.Q\ge 1996$
$d.S\ge 1$
Giải thích các bước giải:
b.Ta có:
$Q=x^2+xy+y^2-3x-3y+1999$
$\to Q=x^2+x(y-3)+y^2-3y+1999$
$\to Q=x^2+2\cdot x\cdot \dfrac{y-3}{2}+(\dfrac{y-3}{2})^2+y^2-3y-(\dfrac{y-3}{2})^2+1999$
$\to Q=(x+\dfrac{y-3}{2})^2+\frac{3y^2-6y-9}{4}+1999$
$\to Q=(x+\dfrac{y-3}{2})^2+\frac{3y^2-6y+3-12}{4}+1999$
$\to Q=(x+\dfrac{y-3}{2})^2+\frac{3(y^2-2y+1)}{4}-3+1999$
$\to Q=(x+\dfrac{y-3}{2})^2+\frac{3(y-1)^2}{4}+1996$
$\to Q\ge 1996$
Dấu = xảy ra khi $y=1,x=1$
d.Ta có:
$S=x^2+26y^2-10xy+14x-76y+59$
$\to S=x^2-2\cdot x\cdot (5y-7)+(5y-7)^2+26y^2-76y-(5y-7)^2+59$
$\to S=(x-5y+7)^2+y^2-6y+10$
$\to S=(x-5y+7)^2+y^2-6y+9+1$
$\to S=(x-5y+7)^2+(y-9)^2+1$
$\to S\ge 1$
Dấu = xảy ra khi $y=9, 38$
