\(M=2x^2+5y^2-2xy+2y+2x=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(4y^2+2y+\dfrac{1}{4}\right)-\dfrac{5}{4}=\left(x-y\right)^2+\left(x+1\right)^2+\left(2y+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\)ta có: (x - y)^2 ≥ 0; (x+1)^2≥ ; (2y+1/2)^2 ≥ 0
=> gtnn M = -5/4
ách nhầm:
\(M=2x^2+5y^2-2xy+2y+2x=\left(x^2+2x+1\right)+\left(x^2-2xy+y^2\right)+4\left(y^2+\dfrac{1}{2}y+\dfrac{1}{16}\right)+\dfrac{3}{4}=\left(x+1\right)^2+\left(x-y\right)^2+4\left(y-\dfrac{1}{4}\right)^2+\dfrac{3}{4}\)
ta có: (x - y)^2 ≥ 0; (x+1)^2≥ ; 4(y+1/4)^2 ≥ 0
vậy gtnn M = 3/4 khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\\\left(y-\dfrac{1}{4}\right)^2=0\end{matrix}\right.\)
!!?
Giải xàm vậy bạn,làm vậy thì dấu "=" khi nào?
giải lại cho nà
M = 2x2 + 5y2 - 2xy + 2y + 2x
M = x2 - 2xy + y2 - 2x + 2y + 1 + x2 + 4x + 4 + 4y2 - 5
M = ( x - y)2 - 2( x - y) + 1 + ( x + 2)2 + 4y2 - 5
M = ( x - y - 1)2 + ( x + 2)2 + 4y2 - 5
Do : ( x - y - 1)2 ≥ 0 ∀x,y
( x + 2)2 ≥ 0 ∀x
4y2 ≥ 0 ∀y
⇒ ( x - y - 1)2 + ( x + 2)2 + 4y2 ≥ 0
⇒ ( x - y - 1)2 + ( x + 2)2 + 4y2 - 5 ≥ - 5
⇒ MMIN = -5 ⇔ x = -2 ; y = -3
