a: \(A=x^2-2x+1+y^2+4y+4+3\)

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

Dấu '=' xảy ra khi x=1 và y=-2

b: \(B=x^2-4x+4+y^2-8y+16-14\)

\(=\left(x-2\right)^2+\left(y-4\right)^2-14>=-14\)

Dấu '=' xảy ra khi x=2 và y=4

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

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

a) \(A=x^2+6x+11\)

\(A=x^2+6x+9+2\)

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

Có: \(\left(x+3\right)^2\ge0\Rightarrow\left(x+3\right)^2+2\ge2\)

Dấu = xảy ra khi: \(\left(x+3\right)^2=0\Rightarrow x+3=0\Rightarrow x=-3\)

Vậy: \(Min_A=2\) tại \(x=-3\)

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

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

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

\(B=5-\left(x-2\right)^2\)

Có: \(\left(x-2\right)^2\ge0\)

\(\Rightarrow5-\left(x-2\right)^2\le5\)

Dấu = xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\)

Vậy: \(Max_B=5\) tại \(x=2\)

d, (x-1) (x+2) (x+3) (x+6)
=(x^2+2x-x-2) (x^2+6x+3x+18)
=(x^2-x^2) + (2x-x+6x-3x) = (-2+18)
=0            + (-8x)              =16
=                    x                =16:(-8)
=                  x                  =-2

A = x^2 + 6x + 11

= x^2 + 6x + 9 + 2

= (x + 3)^2 + 2

min = 2

a, A = x^2 + 6x + 11

= x^2 + 6x + 9 + 2

= (x + 3)^2 + 2

làm tiếp

b, x^2 - 20x + 101

= x^2  20x + 100 + 1

= (x - 10)^2 + 1

có (x - 10)^2 > 0 => (x - 10)^2 +  > 1

Đặt \(A=-2x^2-y^2-2xy+4x+2y+2\)

\(-A=2x^2+y^2+2xy-3x-2y-2\)

\(-A=\left(x^2+2xy+y^2\right)+x^2-4x-2y-2\)

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)-4\)

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

Mà  \(\left(x+y-1\right)^2\ge0\forall x;y\)

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

\(\Rightarrow-A\ge-4\)

\(\Leftrightarrow A\le4\)

Dấu "=" xảy ra khi :

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

Vậy  \(A_{Max}=4\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

Đặt  \(B=x^2-4xy+5y^2+10x-22y+27\)

\(B=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+27\)

\(B=\left[\left(x-2y\right)^2+2\left(x-2y\right)\times5+25\right]+\)\(\left(y^2-2y+1\right)+1\)

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

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

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

\(\Rightarrow B\ge1\)

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  \(B_{Min}=1\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

a/ Ta có:

\(A=x^2-6x+11\)

\(A=x\cdot x-3x-3x+3\cdot3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

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

Vì \(\left(x-3\right)^2\ge0\)

Nên GTNN của \(\left(x-3\right)^2\)là 0

=> \(A_{min}=0+2=2\)

mình chỉ biết a. thôi

a) ta có : \(A=x^2-6x+11\)

\(A=x.x-3x-3x+3.3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

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

vì \(\left(x-3\right)^2\ge0\)

nên GTNN của \(\left(x-3\right)^2\)là \(0\)

\(\Rightarrow\)\(A_{min}\)\(=0+2=2\)

oOo Không đủ can đảm để oOo copy mà nói nhưu mk tự làm