a) (x-2)(x+3)≠ 0 ⇔x ≠-2 và x≠-3
=> ĐKXĐ x ≠-2 và x≠-3
b) \(A=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}\)
=\(\dfrac{x^2-4-5}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2-9}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-3}{x-2}\)
c)|x|=2
th1 x=2 thay vào A ta có
=> \(A=\dfrac{2-3}{2-2}=-\dfrac{1}{1}=-1\)
th2 x=-2 thay vào A ta có
A=\(\dfrac{-2-3}{-2-2}=-\dfrac{5}{-4}=\dfrac{5}{4}\)
để A nguyên thì (x-3) ⋮ (x-2)
vì (x-2) ⋮ (x-2)
=> (x-3)-(x-2) ⋮ (x-2)
<=> (x-3-x+2) ⋮ (x-2)
<=> -1 ⋮ (x-2)
=> (x-2) ∈ Ư(-1)={-1;1}
=> x ∈{1;3}
vậy x ∈{1;3} thì A nguyên