Đáp án:
1.
a, `ĐKXĐ : {x - 3 ne 0`
`{9 - x^2 ne 0`
`{x + 3 ne 0`
`<=> x ne ± 3`
Ta có
`B = 3/(x - 3) - (6x)/(9 - x^2) + x/(x + 3)`
`= 3/(x - 3) + (6x)/[(x - 3)(x + 3)] + x/(x + 3)`
`= [3(x + 3)]/[(x - 3)(x + 3)] + (6x)/[(x - 3)(x + 3)] + [x(x - 3)]/[(x - 3)(x + 3)]`
`= (3x + 9 + 6x + x^2 - 3x)/[(x - 3)(x + 3)]`
`= (x^2 + 6x + 9)/[(x - 3)(x + 3)]`
`= (x + 3)^2/[(x - 3)(x + 3)]`
`= (x + 3)/(x - 3)`
b, `P = A . B = (x + 1)/(x + 3) . (x + 3)/(x - 3) = [(x + 1)(x + 3)]/[(x - 3)(x + 3)] = (x + 1)/(x - 3)`
c, Ta có
`P = (x + 1)/(x - 3) = (x - 3 + 4)/(x - 3) = 1 + 4/(x - 3)`
Để `P ∈ Z <=> 4/(x - 3) ∈ Z`
`<=> x - 3 ∈ Ư(4)`
`<=> x - 3 ∈ {±1 ; ±2 ; ±4}`
`<=> x ∈ {4 ; 2 ; 5 ; 1 ; 7 ; -1}`
2. 1 `ĐKXĐ : x + 1 ne 0 <=> x ne -1`
`(3x^2 + 3x)/(x + 1) = [3x(x + 1)]/(x + 1) = 3x`
2.2
a, `A = (1/(x^2 + x) - (2 - x)/(x + 1)) . (3x)/(1 - 2x + x^2)`
`= (1/[x(x + 1)] - (2 - x)/(x + 1)) . (3x)/(x - 1)^2`
`= (1/[x(x + 1)] - [x(2 - x)]/[x(x + 1)] . (3x)/(x - 1)^2`
`= (1 - x(2 - x))/(x(x + 1)) . (3x)/(x - 1)^2`
`= (1 - 2x + x^2)/(x(x + 1)) . (3x)/(x - 1)^2`
`= (x - 1)^2/(x(x + 1)) . (3x)/(x - 1)^2`
`= [3x(x - 1)^2]/[x(x + 1)(x - 1)^2]`
`= 3/(x + 1)`
b, `Để A ∈ Z <=> 3/(x + 1) ∈ Z`
`<=> x+ 1 ∈ Ư(3)`
`<=> x + 1 ∈ {±1 ; ±3}`
`<=> x ∈ {-2 ; 0 ; 2 ; -4}`
Do `ĐKXĐ : x ne 0`
`<=> x ∈ {-4 ; -2 ; 2}`
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