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) Đk \(x\ne\pm1\), sau khi rút gọn ta được: (bạn tư làm)

   \(P=\frac{x}{x+1}\)

b) Khi \(\left|x-\frac{2}{3}\right|=\frac{1}{3}\) thì hoặc \(x-\frac{2}{3}=\frac{1}{3}\) hoặc \(x-\frac{2}{3}=-\frac{1}{3}\)

Hay là \(x=1\) hoặc \(x=\frac{1}{3}\)

Do để P có nghĩa thì \(x\ne\pm1\) nên \(x=\frac{1}{3}\), khi đó: 

 \(P=\frac{\frac{1}{3}}{\frac{1}{3}+1}=\frac{1}{4}\)

c) P > 1 khi \(\frac{x}{x+1}>1\)

   \(\Leftrightarrow1-\frac{1}{x+1}>1\)

   \(\Leftrightarrow\frac{1}{x+1}< 0\)

   \(\Leftrightarrow x< -1\)

e) Đề không rõ ràng

dễ mà ko bt lm à

ĐKXĐ là gì vậy bạn