Question

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

a, Rút gọn M

b, Tìm x để M>1

c, Tìm x\(\in\)Z để M\(\in\)Z

d, Tìm M khi |x+1|=2

Answer 1

a: Ta có: \(M=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)

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

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

\(=\dfrac{x^2}{x-1}\)

b: Để M>1 thì M-1>0

\(\Leftrightarrow\dfrac{x^2-x+1}{x-1}>0\)

\(\Leftrightarrow x-1>0\)

hay x>1

Answer 2

a) ĐKXĐ: x # 0; x # 1; x# -1

M = (x^2)/(x-1)

Answer 3

b) x > 1