[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: \(x\ne-1;x\ne\pm2\)

`B=(1/(3-sqrtx)-1/(3+sqrtx))*(3+sqrtx)/sqrtx(x>=0,x ne 9)`

`B=((3+sqrtx)/(9-x)-(3-sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=((3+sqrtx-3+sqrtx)/(9-x))*(3+sqrtx)/sqrtx`

`B=(2sqrtx)/((3-sqrtx)(3+sqrtx))*(3+sqrtx)/sqrtx`

`B=2/(3-sqrtx)`

`B>1/2`

`<=>2/(3-sqrtx)-1/2>0`

`<=>(4-3+sqrtx)/[2(3-sqrtx)]>0`

`<=>(sqrtx+1)/(2(3-sqrtx))>0`

Mà `sqrtx+1>=1>0`

`<=>2(3-sqrtx)>0`

`<=>3-sqrtx>0`

`<=>sqrtx<3`

`<=>x<9`

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  \(x\ne1;x\ne-1\)

Ta có :

\(B=\frac{x^2+15}{x^2+3}=\frac{x^2+3+12}{x^2+3}=1+\frac{12}{x^2+3}\)

vì x² \(\ge\)0 \(\Rightarrow\)x² + 3 \(\ge\)3 

\(\Rightarrow\frac{12}{x^2+3}\le4\)

\(\Rightarrow B\le1+4=5\)

Vậy GTLN của B là 5 khi x² + 3 = 3 hay x = 0

Ta có: \(B=1+\frac{12}{x^2+3}\)

Mà \(x^2+3\ne0\in Z\)

\(\Rightarrow\)Ta có 2 trường hợp

+) x²+3 nguyên dương

 \(\Rightarrow\frac{12}{x^2+3}\le12\Rightarrow B\le13\)(1)

+) x²+3 nguyên âm

\(\Rightarrow\frac{12}{x^2+3}< 0\Rightarrow B< 0\)(2)

Từ (1)(2) \(\Rightarrow B\le13\)

\(B=\frac{\left(x^2+15\right)}{x^2+3}\)

\(B=\frac{\left(x^2+3+12\right)}{\left(x^2+3\right)}=1+\frac{12}{\left(x^3+3\right)}\)

B lớn nhất khi \(x=0\Rightarrow B_{MAX}=1+\frac{12}{3}=5\)