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

C = x2 - 4xy + 5y2 + 10x - 22y + 28

= (x^2 - 4xy + 4y^2) + (10x - 20y) + (y^2 - 2y) + 28

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

= (x - 2y)^2 + 2.(x - 2y).5 + 5^2 + (y - 1)^2 + 2

= (x - 2y + 5)^2 + (y - 1)2 + 2

Vì (x−2y+5)^2≥0∀x;y; (y−1)^2≥0∀y nên (x−2y+5)^2+(y−1)^2+2≥2∀x;y

hay C≥2∀x;y

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

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

\(A_{min}=3\) 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 \(\left(x;y\right)=\left(-3;1\right)\)

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

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

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

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

Vậy: \(Min_R=2\Leftrightarrow x=-3;y=1\)

R= x² - 4xy+ 5y² + 10x - 22y + 28

= x² - 2x(2y-5) + (2y-5)² - (2y-5)² +5y² -22y+28

= (x-2y+5)² - 4y² +20y-25 + 5y² -22y +28

= (x-2y+5)² + y² -2y+3

=(x-2y+5)² +(y-1)² +2

Vì (x-2y+5)² ≥0 với mọi x,y

(y-1)² ≥ 0 với mọi y

Suy ra (x-2y+5)² + (y-1)²+2 ≥ 2

Dấu''='' xảy ra <=> x-2y+5=0 và y-1=0

<=> y=1; x=-3

Vậy R _min= 2 ⇔ y=1; x=3

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