Question

Cho M = \(\dfrac{1}{x+1}\) + \(\dfrac{3x-2}{x^3+x^2-x-1}\) và N= \(\dfrac{1}{x-1}+\dfrac{2}{1-x^2}\)

a) Rút gọn M

b) Tính M khi \(\left|2x+1\right|\) = 5

c) Tìm x \(\in\) Z để A = M : N nguyên

Answer 1

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

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

b: |2x+1|=5

=>2x+1=5 hoặc 2x+1=-5

=>2x=4 hoặc 2x=-6

=>x=2(nhận) hoặc x=-3(nhận)

Khi x=2 thì \(M=\dfrac{4+6-3}{\left(2+1\right)\left(2^2-1\right)}=\dfrac{7}{3\cdot3}=\dfrac{7}{9}\)

Khi x=-3 thì \(M=\dfrac{9-9-3}{\left(-3+1\right)\left(9-1\right)}=\dfrac{-3}{\left(-2\right)\cdot8}=\dfrac{3}{16}\)