a) Phân thức M xác định khi và chỉ khi :

+) \(2x-2\ne0\Leftrightarrow x\ne1\)

+) \(2x+2\ne0\Leftrightarrow x\ne-1\)

+) \(1-\frac{x-3}{x+1}\ne0\)

\(\Leftrightarrow x-3\ne x+1\)

\(\Leftrightarrow0x\ne4\left(\text{luôn đúng}\right)\)

Vậy \(x\ne\left\{1;-1\right\}\)

b) \(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)

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

\(M=\left(\frac{2x^2-2x-4-2x^2-4x+6+6x+6}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{4}{x+1}\right)\)

\(M=\frac{8}{2\left(x-1\right)2\left(x+1\right)}\cdot\frac{x+1}{4}\)

\(M=\frac{8\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)\cdot4}\)

\(M=\frac{8\left(x+1\right)}{8\left(x+1\right)\left(x-1\right)}\)

\(M=\frac{1}{x-1}\)

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

\(=\left(\frac{x+1}{2x-2}-\frac{x+3}{2x+2}\right):\left(\frac{4}{x+1}\right)=\left[\frac{\left(x+1\right)\left(2x+2\right)-\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]:\left(\frac{4}{x+1}\right)\)

\(=\left[\frac{2x^2+4x+2-2x^2+2x+6-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)

\(=\left[\frac{6x+8-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)

\(=\frac{14}{4x^2-4}:\left(\frac{4}{x+1}\right)=\frac{14x+14}{16x^2-16}=\frac{7x+7}{8x^2-8}\)

Ta có:

a) M = \(\left(\frac{6x}{x^2-9}-\frac{1}{x+3}+\frac{5}{3-x}\right):\frac{4}{x^2-3x}\)

M = \(\left(\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x^2-3x}{4}\)

M = \(\left(\frac{6x-x+3-5x-15}{\left(x+3\right)\left(x-3\right)}\right)\cdot\frac{x\left(x-3\right)}{4}\)

M = \(\frac{-12.x\left(x-3\right)}{\left(x-3\right)\left(x+3\right).4}\)

M = \(-\frac{3x}{x+3}\)

b) Với x = 2 => M = \(-\frac{3.2}{3+2}=-\frac{6}{5}\)

M = \(\left(\frac{x}{x-3}-\frac{x+3}{3x^2-6x-9}+\frac{1}{3x+3}\right)\)\(\frac{x^2-2x-3}{x^2+x+2}\)

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

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

Ta thấy   x² + x - 2  <   x² + x + 2

nên M < 1