\(M=9x^2+y^2-6x+3y+5\)
\(=\left(9x^2+6x+1\right)+\left(y^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)
\(=\left(3x+1\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Dấu "=" xảy ra khi \(x=-\dfrac{1}{3}\) và \(y=-\dfrac{3}{2}\)
M = 2x2 + y2 - 2xy + 10x - 6y
= (x2 + 4x + 4) + (x2 + 32 + y2 + 6x - 2xy - 6y) - 13
= (x + 2)2 + (x + 3 - y)2 - 13 \(\ge\) - 13
Dấu "=" xảy ra khi x = - 2 và y = 1