a: ĐKXĐ: x<>-2

b: Để M là số nguyên thì \(2x+4-5⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-1;-3;3;-7\right\}\)

Rút gọn:

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

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

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

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

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

<=> M = 

a, Để \(m\) là phân số 

\(2+n\ne0\\ \Rightarrow n\ne-2\)

\(b,\) 

\(\cdot,n=1\\ \Rightarrow m=\dfrac{1-1}{2+1}=\dfrac{0}{3}=0\\ \cdot,n=3\\ \Rightarrow m=\dfrac{1-3}{2+3}=-\dfrac{2}{5}\\ \cdot,n=12\\ \Rightarrow m=\dfrac{1-12}{2+12}=-\dfrac{11}{14}\)

a: ĐKXĐ: n+2<>0

=>n<>-2

b: Sửa đề: m+n=1

m+n=1 thì 1-n=(1-n)/(2+n)

=>(1-n)(2+n)=(1-n)

=>(1-n)(1+n)=0

=>n=1 hoặc n=-1

=>m=0 hoặc m=2

=>m=0 hoặc m=2/1

n=3 thì \(m=\dfrac{1-3}{2+3}=\dfrac{-2}{5}\)

n=12 thì \(m=\dfrac{1-12}{12+2}=\dfrac{-11}{14}\)

ĐK: \(x\ne0;\pm1\)

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

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

b/ Để \(A\in Z\Rightarrow3⋮\left(x+1\right)\Rightarrow x+1=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(x+1=-3\Rightarrow x=-4\)

\(x+1=-1\Rightarrow x=-2\)

\(x+1=1\Rightarrow x=0\left(l\right)\)

\(x+1=3\Rightarrow x=2\)

c/ \(A< 0\Leftrightarrow\dfrac{3}{x+1}< 0\Leftrightarrow x+1< 0\Rightarrow x< -1\)