\(3y^2+x^2+2xy+2x+6y+2017=x^2+2x\left(y+1\right)+\left(y+1\right)^2+\left(2y^2+4y+2\right)+2014\)

\(=\left(x+y+1\right)^2+2\left(y+1\right)^2+2014\ge2014\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}x+y+1=0\\y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\y=-1\end{cases}}\)

Vậy BT đạt GTNN bằng 2014 tại (x;y) = (0;-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

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 $x+y-2=x-1=y-1=0$

$\Leftrightarrow x=y=1$

\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=-2\)

\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)

Dấu \("="\Leftrightarrow x=-5\)

\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)

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

\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(A=x^2+4x+5\)

\(=x^2+4x+4+1\)

\(=\left(x+2\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=-2

\(C=4x^2-4x+5\)

\(=4x^2-4x+1+4\)

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

2A=(2x-y)^2+3(y-2)^2+9>=9

A>=9/2

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

Q = x 2 + 2 y 2 + 2 x y − 2 x − 6 y + 2015        = x 2 + 2 x y + y 2 − 2 x − 2 y + 1 + y 2 − 4 y + 4 + 2010        = x 2 + 2 x y + y 2 − 2 x + 2 y + 1 + y 2 − 4 y + 4 + 2010        = x + y 2 − 2 x + y + 1 + y 2 − 4 y + 4 + 2010        = x + y − 1 2 + y − 2 2 + 2010

A = x² - 10x + 12

= ( x² - 10x + 25 ) - 13

= ( x - 5 )² - 13

( x - 5 )² ≥ 0 ∀ x => ( x - 5 )² - 13 ≥ -13

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

=> MinA = -13 <=> x = 5

B = 6y² + 4y - 1

= 6( y² + 2/3y + 1/9 ) - 5/3

= 6( y + 1/3 )² - 5/3

6( y + 1/3 )² ≥ 0 ∀ x => 6( y + 1/3 )² - 5/3 ≥ -5/3

Đẳng thức xảy ra <=> y + 1/3 = 0 => y = -1/3

=> MinB = -5/3 <=> y = -1/3

C = x² + y² - 2x - 6y - 1

= ( x² - 2x + 1 ) + ( y² - 6y + 9 ) - 11

= ( x - 1 )² + ( y - 3 )² - 11

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y-3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x-1\right)^2+\left(y-3\right)^2-11\ge-11\)

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

=> MinC = -11 <=> x = 1 ; y = 3

D = 2x² + 3y² - x - 3y + 5

= 2( x² - 1/2x + 1/16 ) + 3( y² - y + 1/4 ) + 33/8

= 2( x - 1/4 )² + 3( y - 1/2 )² + 33/8

\(\hept{\begin{cases}2\left(x-\frac{1}{4}\right)^2\ge0\forall x\\3\left(y-\frac{1}{2}\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x-\frac{1}{4}\right)^2+3\left(y-\frac{1}{2}\right)^2+\frac{33}{8}\ge\frac{33}{8}\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{4}=0\\y-\frac{1}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=\frac{1}{2}\end{cases}}\)

=> MinD = 33/8 <=> x = 1/4 ; y = 1/2

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