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

    \(=x^2-4xy+10x+4y^2+25-10y+y^2-2y+3\)

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

Vậy \(GTNN=2\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

C = x² - 4xy + 5y² + 10x - 22y + 28

= (x² - 4xy + 4y²) + (10x - 20y) + (y² - 2y) + 28

= (x - 2y)² + 10(x - 2y) + 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

Vì \(\left(x-2y+5\right)^2\ge0\forall x;y\); \(\left(y-1\right)^2\ge0\forall y\) nên \(\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x;y\)

hay \(C\ge2\forall x;y\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-5\\y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

Vậy ...

hjvbm 

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

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

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

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

Đẳng thức khó tìm quá huhu

\(=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25+\left(y^2-2y+1\right)+2\\ =\left(x-2y\right)^2+10\left(x-2y\right)+25+\left(y-1\right)^2+2\\ =\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

Vậy GTNN của biểu thức là 2

 C = x² - 4xy + 5y² + 10x - 22y + 28

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

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

    = (x - 2y + 5)² + (y - 1)² + 2  \(\ge\)2

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

 C = x² - 4xy + 5y² + 10x - 22y + 28

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

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

    = (x - 2y + 5)² + (y - 1)² + 2  ≥2

Min C = 2 

Thu gọn

D=(x² - 4xy + 4y²) +(y² - 22y + 121) - 93

= (x-2y)² + (y-11)² - 93

Vì (x-2y)² và (y-11)² luôn lớn hơn 0 nên GTNN của biểu thức là -93

Khi đó y=11

và x=22

Ta có

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

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

Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}y=1\\x=-3\end{matrix}\right.\)