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

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

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

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

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

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

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

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

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

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

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

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

a.\(ĐKXĐ:\hept{\begin{cases}x^2-2x\ne0\\x-2\ne0\\x\left(x+1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(x-2\right)\ne0\\x-2\ne0\\x\left(x+1\right)\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-1\end{cases}}}\)

b.\(M=\left(\frac{1}{x^2-2x}+\frac{2}{x-2}\right)\div\frac{2x+1}{x\left(x+1\right)}\)

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

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

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

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

c.Để \(M>1\)thì

 \(\frac{x+1}{x-2}>1\)

c, Ta có : \(M>1\Rightarrow\frac{x+1}{x-2}>1\Leftrightarrow\frac{x+1}{x-2}-1>0\)

\(\Leftrightarrow\frac{x+1-x+2}{x-2}>0\Leftrightarrow\frac{3}{x-2}>0\)

\(\Rightarrow x-2>0\Leftrightarrow x>2\)vì 3 > 0 

d, Để M nguyên khi \(x+1⋮x-2\Leftrightarrow x-2+3⋮x-2\)ĐK : \(x\ne2\)

\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

a) Biểu thức M xác định <=> \(\hept{\begin{cases}2-2x\ne0\\2-2x^2\ne0\end{cases}}\) <=> \(\hept{\begin{cases}2x\ne2\\2x^2\ne2\end{cases}}\) <=> \(\hept{\begin{cases}x\ne1\\x^2\ne1\end{cases}}\) <=> \(\hept{\begin{cases}x\ne1\\x\ne\pm1\end{cases}}\)

Vậy đk xác định biểu thức M <=> x \(\ne\)\(\pm\)1

b) Ta có:

M = \(\frac{x}{2-2x}-\frac{x^2+1}{2-2x^2}\)

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

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

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

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

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

M = \(-\frac{1}{2\left(x+1\right)}\) (đk : x + 1 \(\ne\)0 => x \(\ne\)-1)

Đây là 1 bài trong 1 đề t làm nộp gửi thầy nên t đưa ảnh nha,tại lúc đó đề sai nên trong bài giải có vài chữ ko liên quan

Làm tiếp \(M\ge-3\)

\(\frac{x+1}{2x}\ge-3\)

\(\frac{1}{2}+\frac{1}{2x}\ge-3\)

Đến đây dễ r