Đặt \(A=-2x^2-y^2-2xy+4x+2y+2\)

\(-A=2x^2+y^2+2xy-3x-2y-2\)

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

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)-4\)

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

Mà  \(\left(x+y-1\right)^2\ge0\forall x;y\)

       \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-4\)

\(\Leftrightarrow A\le4\)

Dấu "=" xảy ra khi :

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

Vậy  \(A_{Max}=4\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

Đặt  \(B=x^2-4xy+5y^2+10x-22y+27\)

\(B=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+27\)

\(B=\left[\left(x-2y\right)^2+2\left(x-2y\right)\times5+25\right]+\)\(\left(y^2-2y+1\right)+1\)

\(B=\left(x-2y+5\right)^2+\left(y-1\right)^2+1\)

Mà  \(\left(x-2y+5\right)^2\ge0\forall x;y\)

       \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy  \(B_{Min}=1\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

\(a,A=3x^2-5x+1\)

\(=3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{13}{12}\)

\(=3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\)

Với mọi giá trị của x ta có:

\(\left(x-\dfrac{5}{6}\right)^2\ge0\)

\(\Rightarrow3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)

Vậy Min \(A=-\dfrac{13}{12}\)

Để \(A=-\dfrac{13}{12}\) thì \(x-\dfrac{5}{6}=0\Rightarrow x=\dfrac{5}{6}\)

\(b,B=2x^2+5y^2-4x+2y+4xy+2017\)

\(=\left(2x^2-4x+4xy\right)+5y^2+2y+2017\)

\(=2\left(x^2-2x+2xy\right)+5y^2+2y+2017\)

\(=2\left[x^2-2x\left(1-y\right)+\left(1-y\right)^2\right]+5y^2+2y+2017+2\left(1-y\right)^2\)\(=2\left(x-1+y\right)^2+5y^2+2y+2017-2\left(1-y\right)^2\)

\(=2\left(x+y-1\right)^2+5y^2+2y+2017-2+4y-2y^2\)\(=2\left(x+y-1\right)^2+3y^2+6y+2015\)

\(=2\left(x+y-1\right)^2+3\left(y^2+2y+1\right)+2012\)

\(=2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\)

Với mọi giá trị của x ta có:

\(2\left(x+y-1\right)^2\ge0;3\left(y+1\right)^2\ge0\)

\(\Rightarrow2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\ge2012\) Vậy : Min B = 2012

Để B = 2012 thì \(\left\{{}\begin{matrix}x+y-1=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

A = x² - 4x + 1 = (x² - 2.x.2 + 4) - 3 = (x - 2)² - 3 \(\ge\) -3

Vậy: GTNN của A là -3 (tại x = 2)

B = -2x² + 2x = -2(x² - x) = -2\(\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)\)

= -2\(\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\) \(\le\frac{1}{2}\)

Vậy: GTLN của B là \(\frac{1}{2}\) tại x = \(\frac{1}{2}\)

C = x² + y² + 2x + 2y = (x² + 2x + 1) + (y² + 2y + 1) - 2

= (x + 1)² + (y + 1)² - 2 \(\ge\) -2

Vậy: GTNN của C là -2 tại x = -1 ; y = -1

D = x² - 4xy + 5y² - y = (x² - 4xy + 4y²) + (y² - y + \(\frac{1}{4}\)) - \(\frac{1}{4}\)

= (x - 2y)² + (y - \(\frac{1}{2}\))² - \(\frac{1}{2}\ge-\frac{1}{2}\)

Vậy: GTNN của D là \(\frac{-1}{4}\) tại x = 1 ; y = \(\frac{1}{2}\)

Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn

\(A=4x^2-4xy+5y^2+20x-6y+2044\)

\(=\left(4x^2-4xy+y^2\right)+20x-6y+4y^2+2044\)

\(=\left(2x-y\right)^2+10\left(2x-y\right)+25+\left(4y^2+4y+1\right)+2018\)

\(=\left(2x-y+5\right)^2+\left(2y+1\right)^2+2018\ge2018\)

Dấu "=" xảy ra tại \(y=-\frac{1}{2};x=-\frac{11}{4}\)

Ta có \(A=4x^2-4xy+5y^2+20x-6y+2044\)

            \(=4x^2-4x\left(y-5\right)+\left(y-5\right)^2+4y^2+4y+1+2018\)

            \(=\left(2x-y+5\right)^2+\left(2y+1\right)^2+2018\)

Vì...\(\Rightarrow A\ge2018\)

Dấu bằng xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y+5=0\\2y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{11}{4}\\y=-\frac{1}{2}\end{cases}}}\)

Mấy bạn giải chi tiết ra giùm mình

a: \(A=x^2-2x+1+y^2+4y+4+3\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+3>=3\)

Dấu '=' xảy ra khi x=1 và y=-2

b: \(B=x^2-4x+4+y^2-8y+16-14\)

\(=\left(x-2\right)^2+\left(y-4\right)^2-14>=-14\)

Dấu '=' xảy ra khi x=2 và y=4