\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)

\(A_{min}=3\) khi \(x=-2\)

\(B=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)

\(B_{min}=1\) khi \(x=10\)

\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)

\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(C_{min}=2\) khi \(\left(x;y\right)=\left(-3;1\right)\)

C = ( x² - 4xy + 4y² ) + 10.(x -2y) + ( y² -2y + 1) + 27

   = ( x-2y)² + 2.5.(x-2y) + 25 + (y-1)² + 2

   = ( x-2y + 5 )² + (y-1)² + 2 \(\ge2\)vì \(\left(x-2y+5\right)^2\ge0\forall x,y\) và \(\left(y-1\right)^2\ge0\forall y\)

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

Vậy Min C = 2 \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

\(R=x^2-4xy+5y^2+10x-22y+28\)

\(R=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+28\)

\(R=\left[\left(x-2y\right)^2+2\left(x-2y\right).5+25\right]+\left(y^2-2y+1\right)+2\)

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

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

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

\(\Rightarrow R\ge2\)

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 ...

C = x² - 4xy + 5y² + 10x - 22y + 28

= (x² - 4xy + 4y²) + (10x - 22y) +  25 + y² + 3

= (x - 2y)² + 10(x - 2y) + 25 + y² + 3

= (x - 2y + 5)² + y² + 3 \(\ge\)3

Dấu  " = "  xảy ra  \(\Leftrightarrow\)\(\hept{\begin{cases}x-2y+5=0\\y=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=5\\y=0\end{cases}}\)

Vậy Min  C = 3  \(\Leftrightarrow\)x = 5;  y = 0

M=x²-4xy+5y²+10x-22y+28=(x²+4y²+25-4xy-20y+10x)+(y²-2y+1)+2=(x-2y+5)²+(y-1)²+2

=>M>=2 =>Min M=2

Dấu bằng xảy ra khi:x-2y+5=0 và y-1=0 =>x=-3 và y=1

M=x
2
-4xy+5y
2+10x-22y+28=(x
2+4y
2+25-4xy-20y+10x)+(y
2
-2y+1)+2=(x-2y+5)2+(y-1)2+2
=>M>=2 =>Min M=2
Dấu bằng xảy ra khi:x-2y+5=0 và y-1=0 =>x=-3 và y=1

chúc cậu hok tốt

M=x 2 -4xy+5y 2+10x-22y+28=(x 2+4y 2+25-4xy-20y+10x)+(y 2 -2y+1)+2=(x-2y+5)2+(y-1)2+2

=>M>=2 =>Min M=2

Dấu bằng xảy ra khi:x-2y+5=0 và y-1=0 =>x=-3 và y=1

k cho mk nha

H=\(x^6-2x^3+x^2-2x+2\)

\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)

\(=x^2\left(x^4+2x^3+3x^2+2\right)-2x\left(x^4+2x^3+3x^2+2\right)+\left(x^4+2x^3+3x^2+2\right)\)

\(=\left(x^2-2x+1\right)\left(x^4+2x^3+3x^2+2\right)\)

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

\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)

⇒ MinH=0 ⇔ \(x=1\)

Bài làm:

Ta có: \(x^2-4xy+5y^2+10x-22y+28\)

\(=\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+25+\left(y^2-2y+1\right)+2\)

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

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy Min = 2 khi x = -3 và y = 1

Đặt \(A=x^2-4xy+5y^2+10x-22y+28\)

\(\Rightarrow A=\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+25+\left(y^2-2y+1\right)+2\)

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

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

Vì \(\left(x-2y+5\right)^2\ge0\forall x,y\); \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

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

Vậy \(minA=2\)\(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

x² - 4xy + 5y² + 10x - 22y + 28

= [ ( x² - 4xy + 4y² ) + ( 10x - 20y ) + 25 ] + ( y² - 2y + 1 ) + 2

= [ ( x - 2y )² + 2( x - 2y ).5 + 5² ] + ( y - 1 )² + 2

= ( x - 2y + 5 )² + ( y - 1 )² + 2

\(\hept{\begin{cases}\left(x-2y+5\right)^2\\\left(y-1\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy GTNN của biểu thức = 2 <=> x = -3 ; y = 1