Lời giải:

$M=(x^2+y^2+2xy)+x^2+y^2-6x-6y+11$

$=(x+y)^2+x^2+y^2-6x-6y+11$

$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)+5$

$=(x+y-2)^2+(x-1)^2+(y-1)^2+5\geq 0+0+0+5=5$
Vậy $M_{\min}=5$. Giá trị này đạt tại $x+y-2=x-1=y-1=0$

$\Leftrightarrow x=y=1$

\(Q=x^2+2y^2+2xy-2x-6y+2015\)

\(Q=x^2+2x\left(y-1\right)+2y^2-6y+2015\)

\(Q=x^2+2x\left(y-1\right)+y^2-2y+1+y^2-4y+4+2010\)

\(Q=x^2+2x\left(y-1\right)+\left(y-1\right)^2+\left(y-2\right)^2+2010\)

\(Q=\left(x+y-1\right)^2+\left(y-2\right)^2+2010\ge2010\forall x;y\)

Dấu "=" xảy ra khi x=-3;y=4

2015 nha bạn.

\(Q=x^2+2y^2+2xy-2x-6y+2015\)

\(Q=\left(x^2+y^2+1+2xy-2x-2y\right)+\left(y^2-4y+4\right)+2010\)

\(Q=\left(x+y-1\right)^2+\left(y-2\right)^2+2010\ge2010\)

Dâu'=' xảy ra khi và chỉ khi 

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

Vậy giá trị nhỏ nhất của Q bằng 2010, xảy ra khi x=-1,y=2

\(K=x^2+2y^2-2xy+2x-6y+8\)

\(K=x^2+2x\left(y-1\right)-2y^2-6y+8\)

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

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

\(K=\left(x+y-1\right)^2+\left(y-2\right)^2+4\ge4\forall x;y\)

Dấu "=" xảy ra khi x = -3; y = 4

b: Tham khảo:

a: \(P=x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi x=5/2

Bài làm:

a) \(P=x^2-5x=\left(x^2-5x+\frac{25}{4}\right)-\frac{25}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\le-\frac{25}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(x=\frac{5}{2}\)

Vậy \(Min_P=-\frac{25}{4}\Leftrightarrow x=\frac{5}{2}\)

a) P = x² - 5x 

         = ( x² - 5x + 25/4 ) - 25/4

         = ( x - 5/2 )² - 25/4

( x - 5/2 )² ≥ 0 ∀ x => ( x - 5/2 )² - 25/4 ≥ -25/4

Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2

=> MinF = -25/4 <=> x = 5/2

b) Q = x² + 2y² + 2xy - 2x - 6y + 2015 

         = ( x² + 2xy + y² - 2x - 2y + 1 ) + ( y² - 4y + 4 ) + 2010

         = [ ( x + y )² - 2( x + y ) + 1² ] + ( y - 2 )² + 2010

         = ( x + y - 1 )² + ( y - 2 )² + 2010

\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x,y\\\left(y-2\right)^2\ge0\forall x\end{cases}}\Rightarrow\left(x+y-1\right)^2+\left(y-2\right)^2+2010\ge2010\)

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

=> MinQ = 2010 <=> x = -1 , y = 2

Phần a phải là dấu lớn hơn hoặc bằng nhé, mk đánh nhầm

b) \(Q=x^2+2y^2+2xy-2x-6y+2015\)

\(Q=\left(x^2+2xy+y^2\right)-2\left(x+y\right)+1+\left(y^2-4y+4\right)+2010\)

\(Q=\left(x+y\right)^2-2\left(x+y\right)+1+\left(y-2\right)^2+2010\)

\(Q=\left(x+y-1\right)^2+\left(y-2\right)^2+2010\ge2010\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=-1\\y=2\end{cases}}\)

Vậy \(Min_Q=2010\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}\)

\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right)\\ \)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)\(\ge4\)

Amin=4 khi y=1; x=7

A = (x - y - 6)² - 6y² - 2y - 45 - (y² - 12y - 36)

A = (x - y -6)² + 5(y-1)² +4 \(\ge\)4

Amin = 4 khi y = 1; x = 7

#chanh

\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right) \)

\(A=\left(x-7-6\right)^2+5\left(y-1^2\right)+4\ge4\)

\(Amin=4\)\(khi\)\(y=1;x=7\)

\(A=x^2-2xy-12x+6y^2+2y+45\)

\(=x^2-2x\left(y+6\right)+\left(y+6\right)^2-\left(y+6\right)^2+6y^2+2y+45\)

\(=\left(x-\left(y+6\right)\right)^2-y^2-12y-36+6y^2+2y+45\)

\(=\left(x-y-6\right)^2+5y^2-10y+5+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)

Vậy \(A_{min}=4\)khi \(y=1\)và \(x=7\)

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