~ Bạn tham khảo ~
`a,A = 1/[x-2] + [x^2 - x - 2]/[x^2-7x+10] - [2x-4]/[x-5]`
`=> A = [1(x-5)]/[(x-2)(x-5)] + [(x-2)(x+1)]/[(x-2)(x-5)] - [2(x-2)^2]/[(x-2)(x-5)]`
`=> A = [1(x-5)+(x-2)(x+1)-2(x-2)^2]/[(x-2)(x-5)]`
`=> A = [x-5+(x-2)(x+1-2x+4)]/[(x-2)(x-5)]`
`=> A = [(x-5)+(x-2)(5-x)]/[(x-2)(x-5)]`
`=> A = [(x-5)(1+2-x)]/[(x-2)(x-5)]`
`=> A = [3-x]/[x-2]`
`b,` Ta có :
`A = [3 - x]/[x-2] = -[x-3]/[x-2] = -[x-2-1]/[x-2] = -1+1/[x-2]`
Để `A` nguyên
`=> 1/[x-2]` nguyên
`=> 1 \vdots x-2`
`=> x-2 \in Ư(1)={1;-1}`
`=> x \in {3;1}`
Vậy `x={3;1}`
