Question

Cho A= \(\left(1-\dfrac{4}{x+2}\right):\left(1+\dfrac{1}{x-3}\right)\)

a, Tìm x nguyên để A là số nguyên.

a, Tìm x để A > 0

Answer 1

a) A=(\(\dfrac{x+2}{x+2}\)-\(\dfrac{4}{x+2}\)):(\(\dfrac{x-3}{x-3}\)+\(\dfrac{1}{x-3}\))(ĐKXĐ:x≠-2,x≠3)

⇔A=\(\dfrac{x-2}{x+2}\):\(\dfrac{x-2}{x+3}\)

⇔A=\(\dfrac{x-2}{x+2}\).\(\dfrac{x+3}{x-2}\)

⇔A=\(\dfrac{x+3}{x+2}\)

Answer 2

b)Để A>0

⇔\(\dfrac{x+3}{x+2}\)>0

⇔x+3>0 ,x+2>0 hoặc x+3<0,x+2<0(- với - thành +)

⇔x>-3,x>-2 hoặc x<-3,x<-2

⇔-2<x<-3

Vậy ......