\(B=\left(x-1\right)^2-4\ge4\\ B_{min}=4\Leftrightarrow x=1\)

\(B=x^2-2x-3=\left(x^2-2x+1\right)-4\)

\(=\left(x-1\right)^2-4\ge-4\)

\(minB=-4\Leftrightarrow x=1\)

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

Dấu '=' xảy ra khi x=1

B=\(2x^2-4xy-2x+4y^2+2013\)

\(=x^2-4xy+4y^2+x^2-2x+1+2012\)

\(=\left(x-2y\right)^2+\left(x-1\right)^2+2012\ge2012\)

Dấu = xảy ra khi : \(\left(x-1\right)^2=0\Leftrightarrow x=1\)

                              \(\left(x-2y\right)^2=0\Leftrightarrow2y=1\Leftrightarrow y=\dfrac{1}{2}\)

Vậy \(Min_B=2012\) khi x=1 , y=\(\dfrac{1}{2}\)

\(B=\dfrac{2x^2-12x+25}{x^2-6x+12}=\dfrac{2\left(x^2-6x+12\right)+1}{x^2-6x+12}=2+\dfrac{1}{x^2-6x+9+4}=2+\dfrac{1}{\left(x-3\right)^2+4}\le2+\dfrac{1}{4}=\dfrac{9}{4}\)

Không có min nha bạn . Chỉ có max thôi 

Dấu = xảy ra khi x=3

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

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

\(=5x^2+5\)

Ta thấy \(5x^2\ge0\forall x\)

\(\Rightarrow5x^2+5\ge5\)

\(\Rightarrow B\ge5\)

Dấu "=" xảy ra khi \(x=0\)

...

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

\(=5x^2+5\ge5\)

Dấu "=" xảy ra <=> x^2 = 0 <=> x = 0

GTNN của B là 5 khi x = 0

B=(2x-1)²+(x+2)²

=4x² -4x+x²+4+4x

=5x²+5

xét 5x² ta thấy : 5x² lớn hơn hoặc bằng 0 (với mọi x thuộc R)

=>5x²+5 lớn hơn hoặc bằng 5

=>B lớn hơn hoặc bằng 5

Gía trị nhỏ nhất của B đạt được khi :

B=5 <=> x=0

HOK TỐT NHÉ

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

\(B_{min}=1\) khi \(x=-2\)

\(C=4x^2y^2+12xy+9+6=\left(2xy+3\right)^2+6\ge6\)

\(C_{min}=6\) khi \(xy=-\dfrac{3}{2}\)

Ta có: \(B=2x^2+8x+9\)

\(=2\left(x^2+4x+\dfrac{9}{2}\right)\)

\(=2\left(x^2+4x+4+\dfrac{1}{2}\right)\)

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

Dấu '=' xảy ra khi x=-2

Vậy: \(B_{min}=1\) khi x=-2

[2x-2=0=>x=1

x-1=0=>x=1

x+1=0=>x=-1

5=0=>x=5

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)

\(=\left(\dfrac{x-1}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(\dfrac{x+1-x-3}{x+1}\right)\)

\(=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{-2}{x+1}\)

\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}\)

\(=\dfrac{-2x+2}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)

\(=\dfrac{-2\left(x-1\right)}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)

\(=\dfrac{1}{2}\)

Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)

a/

Để biểu thức được xác định

\(=>\left\{{}\begin{matrix}2x-2\ne0\\2x+2\ne0\\x+1\ne0\end{matrix}\right.\)

\(\odot2x-2\ne0\)

\(2x\ne2\)

\(x\ne1\)

\(\odot2x+2\ne0\)

\(2x\ne-2\)

\(x\ne-1\)

\(\odot x+1\ne0\)

\(x\ne-1\)

Vậy điều kiện xác định của bt là: \(x\ne-1;x\ne\pm2\)

a)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(M=x^4-x^3-x^3+x^2+x^2-2x+1\)

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

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

\(=\left(x-1\right)^2\cdot x^2+\left(x-1\right)^2=\left(x-1\right)^2\left(x^2+1\right)\)

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

\(x^2+1\ge1\)\(\forall x\)

Do đó: \(M>=1\)

Dấu = xảy ra khi x=0

a)  ĐK : \(x\ne1\); \(x\ne-1\)

b) Ta có biểu thức:

\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\left(\frac{4x^2-4}{5}\right)\)

\(=\left(\frac{x+1}{2.\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}-\frac{x+3}{2.\left(x+1\right)}\right).\left(\frac{4.\left(x^2-1\right)}{5}\right)\)

\(=\frac{\left(x+1\right)^2+3.2-\left(x+3\right)\left(x-1\right)}{2.\left(x-1\right)\left(x+1\right)}.\frac{4.\left(x+1\right)\left(x-1\right)}{5}\)

\(=\frac{x^2+2x+2+6-x^2-2x+3}{2.\left(x-1\right)\left(x+1\right)}.\frac{4.\left(x+1\right)\left(x-1\right)}{5}=\frac{40.\left(x+1\right)\left(x-1\right)}{10.\left(x+1\right)\left(x-1\right)}=4\)

Vậy giá trị của biểu thức B không phụ thuộc vào biến x khi  \(x\ne1;x\ne-1\)