A = 2x2 + 5y2 - 2xy + 2x + 2y
A = x2 - 2xy + y2 - 2x - 2y + 1 + x2 + 4x + 4 + 4y2 + 4y + 1 - 6
A = ( x - y)2 - 2( x + y) + 1 + ( x + 2)2 + ( 2y + 1)2 - 6
A = ( x - y - 1)2 + ( x + 2)2 + ( 2y + 1)2 - 6
Do : ( x - y - 1)2 ≥ 0 ∀x
( x + 2)2 ≥ 0 ∀x
( 2y + 1)2 ≥ 0 ∀x
⇒ ( x - y - 1)2 + ( x + 2)2 + ( 2y + 1)2 ≥ 0
⇒ ( x - y - 1)2 + ( x + 2)2 + ( 2y + 1)2 - 6 ≥ -6
⇒ AMIN = - 6
Ah , sorry bạn nha , mk làm nhầm rùi
A = 2x2 + 5y2 - 2xy + 2x + 2y
A = x2 - 4xy + 4y2 + x2 + 2xy + y2 + 2x + 2y + 1 - 1
A = ( x - 2y)2 + ( x + y)2 + 2( x + y) + 1 - 1
A = ( x - 2y)2 + ( x + y + 1)2 - 1
Do : ( x - 2y)2 ≥ 0 ∀x
( x + y + 1)2 ≥ 0 ∀x
⇒ ( x - 2y)2 + ( x + y + 1)2 ≥ 0
⇒ ( x - 2y)2 + ( x + y + 1)2 - 1 ≥ - 1
⇒ AMIN = -1 ⇔ x = \(\dfrac{-1}{3};y=\dfrac{-2}{3}\)
Đóng góp cách khác :))
\(A=2x^2+5y^2-2xy+2x+2y\)
\(2A=4x^2+10y^2-4xy+4x+4y\)
\(2A=\left(4x^2-4xy+y^2\right)+4x-2y+1+9y^2+6y+1-2\)
\(2A=\left(2x-y\right)^2+2\left(2x-y\right)+1+\left(3y+1\right)^2-2\)
\(2A=\left(2x-y+1\right)^2+\left(3y+1\right)^2-2\ge-2\)
\(\Rightarrow A\ge-1\)
Dấu"=" xảy ra khi \(\left\{{}\begin{matrix}2x-y+1=0\\3y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
