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

\(A=2x^2+y^2+2xy-6x-2y+10\)

<=>\(A=y^2+2y\left(x-1\right)+2x^2-6x+10\)

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

<=>\(A=y^2+2y\left(x-1\right)+\left(x-1\right)^2+\left(x-2\right)^2+5\)

<=>\(A=\left(y+x-1\right)^2+\left(x-2\right)^2+5\ge5\)

=> A đạt giá trị nhỏ nhất là 5 khi \(\hept{\begin{cases}\left(y+x-1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y+x-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}\)

\(S=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(y^2-4y+4\right)+2021\)

\(S=\left(x+y+1\right)^2+\left(y-2\right)^2+2021\ge2021\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(-3;2\right)\)

x^2+y^2+2x+2y+2xy+5

bài này rễ quá

giá trị nhỏ nhất của A là 5 nhé

CMR B \(\ge\)một số nào đó

\(B=2x^2+2xy+y^2-2x+2y+2016\)

\(=\left(x^2+2xy+y^2+2x+2y+1\right)+\left(x^2-4x+4\right)+2011\)

\(=\left[\left(x+y\right)^2+2\left(x+y\right)+1\right]+\left(x-2\right)^2+2011\)

\(=\left(x+y+1\right)^2+\left(x-2\right)^2+2011\ge2011\forall x;y\)có GTNN là 2011

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy \(B_{min}=2011\) tại \(x=2;y=-3\)

thang ngu

a) \(A=4x^2-12x+100=\left(2x\right)^2-12x+3^2+91=\left(2x-3\right)^2+91\)

Ta có: \(\left(2x-3\right)^2\ge0\forall x\inℤ\)

\(\Rightarrow\left(2x-3\right)^2+91\ge91\)

hay A \(\ge91\)

Dấu "=" xảy ra <=> \(\left(2x-3\right)^2=0\)

<=> 2x-3=0

<=> 2x=3

<=> \(x=\frac{3}{2}\)

Vậy Min A=91 đạt được khi \(x=\frac{3}{2}\)

b) \(B=-x^2-x+1=-\left(x^2+x-1\right)=-\left(x^2+x+\frac{1}{4}-\frac{5}{4}\right)=-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)

Ta có: \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\) hay B\(\le\frac{5}{4}\)

Dấu "=" \(\Leftrightarrow-\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy Max B=\(\frac{5}{4}\)đạt được khi \(x=\frac{-1}{2}\)

\(C=2x^2+2xy+y^2-2x+2y+2\)

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

\(\Leftrightarrow C=\left(x+y-1\right)^2+x^2+1\)

Ta có: 

\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x;y\inℤ\\x^2\ge0\forall x\inℤ\end{cases}}\)

\(\Leftrightarrow\left(x+y-1\right)^2+x^2+1\ge1\)

hay C\(\ge\)1

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x+y-1\right)^2=0\\x^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=1\\x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=0\end{cases}}}\)

Vậy Min C=1 đạt được khi y=1 và x=0

Ta có : \(x^2+y^2-2x+4y+1\)

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

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

Vì \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\in R\)

Nên : \(A=\left(x-1\right)^2+\left(y+2\right)^2-4\ge-4\forall x,y\in R\)

Vậy \(A_{min}=-4\) khi x = 1 và y = -2

hoc tot de lam lien doi nho chua.

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

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

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

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

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

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra khi :

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

Vậy Min A = 2 khi x=y=1

\(B=x^2-2xy+2y^2+2x-10y+17\)

\(B=\left(x^2-2xy+y^2\right)+y^2+2x-10y+17\)

\(B=\left[\left(x-y\right)^2+2\left(x-y\right)+1\right]-8y+y^2+16\)

\(B=\left(x-y+1\right)^2+\left(y^2-8y+16\right)\)

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

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

       \(\left(y-4\right)^2\ge0\forall y\)

\(\Rightarrow B\ge0\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x-y+1=0\\y-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=4\end{cases}}\)

Vậy Min B = 0 khi (x;y)=(3;4)