`M = 8x^2 + 20y^2 + 20xy + 6x − 7`
`= 3(x^2 + 2x + 1) + 5(x^2 + 4y^2 + 4xy) − 10`
`= 3(x + 1)^2 + 5(x + 2y)^2 − 10`
Ta có :
`(x + 1)^2 ≥ 0⇒ 3(x + 1)^2≥0`
`(x + 2y)^2 ≥ 0⇒5(x + 2y)^2≥0`
`⇒ 3(x + 1)^2 + 5(x + 2y)^2 − 10≥ -10 ∀x,y ∈ R`
Dấu "=" xảy ra khi:
`+)x + 1 = 0 ⇔ x=-1`
`+)x + 2y = 0⇔-1+2y=0⇔y =1/2`
Vậy `M_(min)=-10 ⇔x=-1;y=1/2`