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\(A=\left(x^2+y^2+36-2xy-12x+12y\right)+5y^2-10y+5+109\)
\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+109\ge109\)
\(A_{min}=109\) khi \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: B = x2 + 2y2 - 2xy + 2x - 6y + 10
B = (x2 - 2xy + y2) + 2x - 6y + y2 + 10
B = (x - y)2 + 2(x - y) + 1 - 4y + y2 + 4 + 5
B = (x - y + 1)2 + (y - 2)2 + 5 \(\ge\)5 \(\forall\)x;y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y+1=0\\y-2=0\end{cases}}\) <=> \(\hept{\begin{cases}x=y-1\\y=2\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy MinB = 5 <=> x = 1 và y = 2
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\(H=x^2+2y^2-2xy+6y+2023\\=(x^2-2xy+y^2)+(y^2+6y+9)+2014\\=(x-y)^2+(y^2+2\cdot y\cdot3+3^2)+2014\\=(x-y)^2+(y+3)^2+2014\)
Ta thấy: \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(y+3\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-y\right)^2+\left(y+3\right)^2\ge0\forall x;y\)
\(\Rightarrow H=\left(x-y\right)^2+\left(y+3\right)^2+2014\ge2014\forall x;y\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-3\end{matrix}\right.\)
\(\Leftrightarrow x=y=-3\)
Vậy \(Min_H=2014\) khi \(x=y=-3\)
\(H=x^2+2y^2-2xy+6y+2023\)
\(2H=2x^2+4y^2-4xy+12y+4046\)
\(2H=4y^2-4y\left(x-3\right)+\left(x-3\right)^2-\left(x-3\right)^2+2x^2+4046\)
\(2H=\left(2y-x+3\right)^2+x^2+6x+9+4028\)
\(H=\dfrac{1}{2}\left[\left(2y-x+3\right)^2+\left(x+3\right)^2\right]+2014\)
Vì \(\left(2y-x+3\right)^2+\left(x+3\right)^2\ge0\forall x,y\)
\(MinH=2014\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3\end{matrix}\right.\)
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Lời giải:
$M=(x^2+y^2+2xy)+x^2+y^2-6x-6y+11$
$=(x+y)^2+x^2+y^2-6x-6y+11$
$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+5$
$=(x+y-2)^2+(x-1)^2+(y-1)^2+5\geq 0+0+0+5=5$
Vậy $M_{\min}=5$. Giá trị này đạt tại $x+y-2=x-1=y-1=0$
$\Leftrightarrow x=y=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow x=-2\)
\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)
Dấu \("="\Leftrightarrow x=-5\)
\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(A=x^2+4x+5\)
\(=x^2+4x+4+1\)
\(=\left(x+2\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=-2
\(C=4x^2-4x+5\)
\(=4x^2-4x+1+4\)
\(=\left(2x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right)\\ \)
\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)\(\ge4\)
Amin=4 khi y=1; x=7
\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right) \)
\(A=\left(x-7-6\right)^2+5\left(y-1^2\right)+4\ge4\)
\(Amin=4\)\(khi\)\(y=1;x=7\)
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Câu 1 :
\(E=4x^2+y^2-4x-2y+3\)
\(E=\left(2x\right)^2-2\cdot2x\cdot1+1^2+y^2-2\cdot y\cdot1+1^2+1\)
\(E=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=1\end{cases}}\)
Câu 2 :
\(G=x^2+2y^2+2xy-2y\)
\(G=x^2+2xy+y^2+y^2-2.y\cdot1+1^2-1\)
\(G=\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\y=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Do A nhỏ nhất
Suy ra : x^2 = 0, 2y^2 = 0 , 4y = 0 .......( tất cả số hạng bằng 0)
Suy ra A= 2019
\(A=x^2+2y^2+4y+2xy-4x+2019\)
\(A=\left(x^2+y^2-2^2+2xy-4y-4x\right)+\left(y^2+8y+4^2\right)+2007\)
\(A=\left(x+y-2\right)^2+\left(y+4\right)^2+2007\ge2007\)
Vậy \(Min_A=2007\) khi \(\hept{\begin{cases}x+y-2=0\\y+4=0\end{cases}}\hept{\begin{cases}x+y=2\\y=-4\end{cases}}\hept{\begin{cases}x=6\\y=4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2-2xy+6y^2-12x+2y+54\)
\(A=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5+4\)
\(A=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y^2-2y+1\right)+4\)
\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)
Do: \(\left(x-y-6\right)^2\ge0\forall xy\); \(5\left(y-1\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-y-6\right)^2+5\left(y-1\right)^2\ge0\)
\(\Leftrightarrow A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)
\(\Rightarrow A_{Min}=4\)
Dấu "=" xảy ra khi \(x=7;y=1\)
<=> x^2 + 2x(y+2) + y^2+4y+4+y^2+2y+1-4
<=> x^2 + 2x(y+2) + (y+2)^2 + (y+1)^2 - 4
<=> (x+y+2)^2 + (y+1)^2 - 4 >= -4
min = -4 khi y = -1 , x = -1
\(=\left(x+y+2\right)^2+\left(y+1\right)^2-4\)
Vì \(\left(x+y+2\right)^2\ge0\forall x\) , \(\left(y+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+y+2\right)^2+\left(y+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+y+2\right)^2+\left(y+1\right)^2-4\ge-4\forall x\)
Vậy GTNN của A=-4 Dấu bằng xảy ra khi
\(\Rightarrow\hept{\begin{cases}\left(x+y+2\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2-y\\y=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=-1\end{cases}}\)
Vậy GTNN của A=-4 khi và chỉ khi x=-3 , y=-1