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\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
D=\(\left(x^2-2xy+y^2+4x-4y+4\right)+\left(x^2-2x+1\right)+4\)\(=\left(x+y+2\right)^2+\left(x+1\right)^2+4\ge4\Rightarrow Min_D=4\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x+y=-2\Rightarrow y=-1\end{matrix}\right.\)
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H=2x2+y2+2xy+6x+4y+9
= (x2+2xy+y2)+(4x+4y)+4+(x2+2x+1)+4
= (x+y)2+4(x+y)+4 +(x+1)2+4
=(x+y+2)2 +(x+1)2+4
=> MinH =4 khi x=-1; y=-1
H=2x2+y2+2xy+6x+4y+9
= (x2+2xy+y2)+(4x+4y)+4+(x2+2x+1)+4
= [(x+y)2+4(x+y)+4] +(x+1)2+4
=(x+y+2)2 +(x+1)2+4
do (x+y+2)2≥0 ∀x;y
(x+1)2 ≥0 ∀x
=> (x+y+2)2 +(x+1)2+4 ≥4
=> min H= 4 khi x=-1;y=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(G=2x^2+2y^2+z^2+2xy-2xz-2yz-2x-4y\)
\(=\left[x^2+2x\left(y-z\right)+\left(y-z\right)^2\right]+\left(x^2-2x+1\right)+\left(y^2-4y+4\right)-5\)
\(=\left(x+y-z\right)^2+\left(x-1\right)^2+\left(y-2\right)^2-5\ge-5\)
\(minG=-5\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y-z=0\\x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\\z=3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Đặt `A=2x^2+2y^2+2xy-4x+4y+2021`
`<=>2A=4x^2+4y^2+4xy-8x+8y+4042`
`<=>2A=4x^2+4xy+y^2-8x-4y+3y^2+12y+4042`
`<=>2A=(2x+y)^2-4(2x+y)+4+3y^2+12y+12+4026`
`<=>2A=(2x+y-2)^2+3(y+2)^2+4026>=4026`
`=>A>=2013`
Dấu "=" xảy ra khi `y=-2,x=(2-y)/2=2`
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=2x^2+y^2-2xy+2x-4y+9\)
\(=\left(x^2-2xy+y^2\right)+2.\left(x-y\right).2+4+x^2-2x+5\)
\(=\left(x-y\right)^2+2.\left(x-y\right).2+2^2+\left(x^2-2x+1\right)+4\)
\(=\left(x-y+2\right)^2+\left(x-1\right)^2+4\)
Ta thấy : \(\hept{\begin{cases}\left(x-y+2\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}\forall x,y}\) \(\Rightarrow\left(x-y+2\right)^2+\left(x-1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-y+2\right)^2+\left(x-1\right)^2+4\ge4\forall x,y\)
hay : \(A\ge4\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y+2\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}y=3\\x=1\end{cases}}\)
Vậy : min \(A=4\) tại \(\hept{\begin{cases}y=3\\x=1\end{cases}}\)
Tìm GTNN chứ k phải tìm x,y bn ơi