( (2x-1)/(x+3) - x/(3-x) - (3-10x)/(x^2 -9) ) : (x+2) / (x-3)  (x khác ±3, x khác -2)

= ( (2x-1)/(x+3) + x/(x-3) - (3-10x)/((x-3)(x+3)) ) : (x+2) / (x-3)

= [ (2x-1)(x-3) + x(x+3) - (3-10x) ] / ((x-3)(x+3)) : (x+2) / (x-3)

= [ (2x^2 - 7x + 3) + (x^2 + 3x) - (3 - 10x) ] / ((x-3)(x+3)) : (x+2) / (x-3)

= [ 3x^2 + 6x ] / ((x-3)(x+3)) : (x+2) / (x-3)

= [ 3x(x + 2) ] / ((x-3)(x+3)) : (x+2) / (x-3)

= [ 3x(x + 2) ] / ((x-3)(x+3)) * (x-3) / (x+2)

= [ 3x(x + 2)(x-3) ] / [ (x-3)(x+3)(x+2) ]

= 3x / (x+3)

Để 3x / (x+3) có giá trị nguyên, thì 3 - 9 / (x+3) phải nguyên

=> 9 / (x+3) phải nguyên

=> (x+3) thuộc Ư(9) = {±1, ±3, ±9}

Suy ra:

x + 3 = 1 => x = -2 (Loại)

x + 3 = -1 => x = -4 (tm)

x + 3 = 3 => x = 0 (tm)

x + 3 = -3 => x = -6 (tm)

x + 3 = 9 => x = 6 (tm)

x + 3 = -9 => x = -12 (tm)