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\(A=2x^2+5y^2-2xy+2x+2y\)
\(=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(4y^2+2.2y.\frac{1}{2}+\frac{1}{4}\right)-1-\frac{1}{4}\)
\(=\left(x-y\right)^2+\left(x+1\right)^2+\left(2y+\frac{1}{2}\right)^2-\frac{5}{4}\)
Ta thấy: \(\left(x-y\right)^2\ge0;\left(x+1\right)^2\ge0;\left(2y+\frac{1}{2}\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(x-y\right)^2+\left(x+1\right)^2+\left(2y+\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
\(\Rightarrow Min_A=-\frac{5}{4}\).
\(M=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+3y^2-2\)
\(M=\left(x-y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2+3y^2-2\ge-2\)
A=2x2+5y2 -2xy+2x+2y
<=> A=(x2+2xy+y2)+(2x+2y)+1+(x2-4xy+4y2)-1
<=> A=(x+y)2 +2(x+y)+1 +(x-2y)2 -1
<=> A=(x+y+1)2 +(x-2y)2-1
=> GTNN của A=-1 dấu = xảy ra khi x=\(-\dfrac{2}{3}\) ;y=\(\dfrac{-1}{3}\)
a/ A = 2x2 + y2 - 2xy - 2x + 3
= (x2 - 2xy + y2) + (x2 - 2x + 1) + 2
= (x - y)2 + (x - 1)2 + 2\(\ge2\)
\(2xy+2x-5z=0\Rightarrow5z=2xy+2x\Rightarrow z=\frac{2}{5}xy+\frac{2}{5}x\)
\(A=x^2+2y^2+2xy+\frac{8}{5}y+z+2\)
\(A=x^2+2y^2+2xy+\frac{8}{5}y+\frac{2}{5}xy+\frac{2}{5}x+2\)
\(A=x^2+2y^2+\frac{12}{5}xy+\frac{2}{5}x+\frac{8}{5}y+2\)
\(A=x^2+\left(\frac{6y}{5}\right)^2+\left(\frac{1}{5}\right)^2+2.\frac{6}{5}xy+\frac{2}{5}x+\frac{12y}{25}+\frac{14}{25}y^2+\frac{28y}{25}+\frac{14}{25}+\frac{7}{5}\)
\(A=\left(x+\frac{6y}{5}+\frac{1}{5}\right)^2+\frac{14}{25}\left(y+1\right)^2+\frac{7}{5}\ge\frac{7}{5}\)
\(\Rightarrow A_{min}=\frac{7}{5}\) khi \(\left\{{}\begin{matrix}x=1\\y=-1\\z=0\end{matrix}\right.\)
\(A=2x^2+5y^2-2xy+2y+2x\)
\(=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(x^2-4xy+4y^2\right)-1\)
\(=\left(x+y+1\right)^2+\left(x-2y\right)^2-1\)
Ta thấy :(x + y +1)2 ≥ 0 ∀ x,y
(x - 2y)2 ≥ 0 ∀ x,y
⇒ (x + y +1)2 +(x - 2y)2 ≥ 0 ∀ x,y
⇔(x + y +1)2 +(x - 2y)2 -1 ≥ -1 ∀ x,y
⇔ A ≥ -1 ∀ x,y
Vậy GTNN của A là -1 \(\Leftrightarrow\left\{{}\begin{matrix}x+y+1=0\\x-2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x=2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{-1}{3}\end{matrix}\right.\)