\(A=x^2+2y^2+2xy-4x+6y+2020\)

\(A=\left(x^2+y^2+2^2+2xy-4y-4x\right)+\left(y^2+10y+25\right)+1991\)

\(A=\left(x+y-2\right)^2+\left(y+5\right)^2+1991\ge1991\)

Vậy \(Min_A=1991\)khi \(\hept{\begin{cases}x+y-2=0\\y+5=0\end{cases}}\hept{\begin{cases}x+y=2\\y=-5\end{cases}}\hept{\begin{cases}x=7\\y=-5\end{cases}}\)

Ai giúp mik với ạ 😢😭😭😭😭😢😷

\(A=\left(x^2+y^2+36-2xy-12x+12y\right)+5y^2-10y+5+109\)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+109\ge109\)

\(A_{min}=109\) khi \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

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

\(H=x^2+2y^2-2xy+6y+2023\\=(x^2-2xy+y^2)+(y^2+6y+9)+2014\\=(x-y)^2+(y^2+2\cdot y\cdot3+3^2)+2014\\=(x-y)^2+(y+3)^2+2014\)

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 \(\left(2y-x+3\right)^2+\left(x+3\right)^2\ge0\forall x,y\)

\(MinH=2014\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3\end{matrix}\right.\)

x^2 - 2xy + 6y^2 - 12x + 2y +45 
= x^2 - 2x(y+6) + (y+6)^2 - (y+6)^2 + 6y^2 +2y + 45 
= (x - y - 6)^2 - y^2 - 12y - 36 + 6y^2 + 2y + 45 
= (x - y - 6)^2 + 5y^2 - 10y + 9 
= (x - y - 6)^2 + 5.(y^2 - 2y +1) + 4 
= (x - y - 6)^2 + 5.(y-1)^2 + 4 
=>> MIN = 4 khi (x;y) = {(7;1)}

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(=x^2+y^2+36-2xy-12x+12y+5y^2-10y+5+4\)

\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

GTNN A = 4 Khi: \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}\Rightarrow\hept{\begin{cases}y=1\\x=7\end{cases}}}\)

\(A=x^2-2xy+6y^2-12x+2y\)\(+45\)

 \(=x^2+y^2+36-2xy-12x\)\(+12y+5y^2-10y+5+4\)

 \(=\left(x-y-6\right)^2+5\left(y-1\right)^2\)\(+4\ge4\)

GTNN của A là 4 khi \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}y=1\\x-y=6\end{cases}\Rightarrow\hept{\begin{cases}y=1\\x=7\end{cases}}}\)

Vậy BT A đạt giá trị nhỏ nhất là 4 tại x = 7 và y = 1

\(Q=x^2+2y^2+2xy-2x-6y+2015\)

\(Q=x^2+2x\left(y-1\right)+2y^2-6y+2015\)

\(Q=x^2+2x\left(y-1\right)+y^2-2y+1+y^2-4y+4+2010\)

\(Q=x^2+2x\left(y-1\right)+\left(y-1\right)^2+\left(y-2\right)^2+2010\)

\(Q=\left(x+y-1\right)^2+\left(y-2\right)^2+2010\ge2010\forall x;y\)

Dấu "=" xảy ra khi x=-3;y=4

2015 nha bạn.

\(Q=x^2+2y^2+2xy-2x-6y+2015\)

\(Q=\left(x^2+y^2+1+2xy-2x-2y\right)+\left(y^2-4y+4\right)+2010\)

\(Q=\left(x+y-1\right)^2+\left(y-2\right)^2+2010\ge2010\)

Dâu'=' xảy ra khi và chỉ khi 

\(\hept{\begin{cases}x+y-1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)

Vậy giá trị nhỏ nhất của Q bằng 2010, xảy ra khi x=-1,y=2

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 $x+y-2=x-1=y-1=0$

$\Leftrightarrow x=y=1$

\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)

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 \(x=\dfrac{1}{2}\)

\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right)\\ \)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)\(\ge4\)

Amin=4 khi y=1; x=7

A = (x - y - 6)² - 6y² - 2y - 45 - (y² - 12y - 36)

A = (x - y -6)² + 5(y-1)² +4 \(\ge\)4

Amin = 4 khi y = 1; x = 7

#chanh

\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right) \)

\(A=\left(x-7-6\right)^2+5\left(y-1^2\right)+4\ge4\)

\(Amin=4\)\(khi\)\(y=1;x=7\)