Question

Answer 1

a) A= \(\dfrac{x+5}{x+2}\)
   A = \(\dfrac{\left(x+2\right)+3}{x+2}\)
   A = \(1+\dfrac{3}{x+2}\)
Để A là số nguyên

=> \(\dfrac{3}{x+2}\) thuộc số nguyên
=>x+2 ∈ Ư(3)={1;-1;3;-3)

x+2	-1	1	3	-3
x	-3	-1	1	-5

Vậy x ∈ {-1;-3;1;-5} thì A nguyên

b)  A= \(\dfrac{x+5}{x+2}\)
A = \(\dfrac{\left(x+2\right)+3}{x+2}\)
   A = \(1+\dfrac{3}{x+2}\)

Để A có GTLN thì \(\dfrac{3}{x+2}\) ∈ Ư(3) = {1,-1,3,-3}

Ta có

\(\dfrac{3}{1+2}\)  = \(\dfrac{3}{3}\) = 1

\(\dfrac{3}{-1+2}\) = \(\dfrac{3}{1}\)= 3

\(\dfrac{3}{3+2}\) = \(\dfrac{3}{5}\)

\(\dfrac{3}{-3+2}\) = \(\dfrac{3}{-1}\) = -3

=> A = -1 thì A là GTLN