Show that 1/(5x^2-13x-6) - 1/(x^2 -9= Ax + B/(x-3)(x+3)(5x+2) and find the value of A and B?

1/(5x^2-13x-6) - 1/(x^2 -9)
1/(5x+2)(x-3) -1/(x+3)(x-3)
(x+3)/(x-3)(x+3)(5x+2) -(5x+2)/(x-3)(x+3)(5x+2)
x+3-5x-2/(x-3)(x+3)(5x+2)
-4x+1/(x-3)(x+3)(5x+2)=Ax + B/(x-3)(x+3)(5x+2)
Ax=-4x------>A=-4
B=+1

1/(5x^2 - 13x - 6) - 1/(x^2 - 9)
= (x^2 - 9) - (5x^2 - 13x - 6) / (5x^2 - 13x - 6)(x^2 - 9)
= (-4x^2 + 13x - 3) / (x - 3)^2(x + 3)(5x + 2)
= -(4x - 1)(x - 3) / (x - 3)^2(x + 3)(5x + 2)
= -(4x - 1) / (x - 3)(x + 3)(5x + 2)

1/(5x^2 - 13x - 6) - 1/(x^2 - 9)
= Ax + B/(x - 3)(x + 3)(5x + 2)
= -4x + 1 / (x - 3)(x + 3)(5x + 2)
Solution:
A = -4, B = 1

I take it you mean (Ax + B)/((5x + 2)(x + 3)(x - 3)). Parentheses!!

1/((5x + 2)(x - 3)) - 1/((x + 3)(x - 3)) =

(x + 3)/((5x + 2)(x + 3)(x - 3)) - (5x + 2)/((5x + 2)(x + 3)(x - 3)) =

(-4x + 2)/((5x + 2)(x + 3)(x - 3))

5x^2-13x-6
= 5x^2-15x+2x-6
= 5x(x-3) + 2(x-3)
= (5x+2)(x-3)

1/(5x^2-13x-6) - 1/(x^2 -9 )= Ax + B/(x-3)(x+3)(5x+2)
1/ ((5x+2)(x-3)) - 1/((x-3)(x+3)) = Ax + B / ((x-3)(x+3)(5x+2))
Multiply both sides by (x-3)(x+3)(5x+2)

(x+3) - (5x+2) = Ax (x-3)(x+3)(5x+2) + B
x+3-5x-2 = Ax (x-3)(x+3)(5x+2) + B
-4x+1 = Ax (x-3)(x+3)(5x+2) + B

Match the coefficient of x
-4 = -18A
A = 2/9

Match the constant terms
1 = B