Answer:
[tex]\boxed{\sf \ \ \ \dfrac{18}{(x-7)(x+12)(x+30)} \ \ \ }[/tex]
Step-by-step explanation:
hello,
first of all we will study the quadratic expressions
we can write that, (the different answers provide good clues :-) )
[tex]x^2+5x-84=(x-7)(x+12)[/tex]
and
[tex]x^2+23x-210=(x-7)(x+30)[/tex]
so first of all as we cannot divide by 0 we need to take x different from 7, -12 and -30 and then we can write
[tex]\dfrac{1}{x^2+5x-48}-\dfrac{1}{x^2+23x-210}=\dfrac{1}{(x-7)(x+12)}-\dfrac{1}{(x-7)(x+30)}\\\\=\dfrac{(x+30) -(x+12)}{(x-7)(x+12)(x+30)}=\dfrac{x+30-x-12}{(x-7)(x+12)(x+30)}\\\\=\dfrac{18}{(x-7)(x+12)(x+30)}[/tex]
hope this helps
