Đáp án:
1.
a, `ĐKXĐ : x^2 - 4 ne 0 <=> x ne +- 2`
Ta có
`A = (x^2 - 4x + 4)/(x^2 - 4) = (x - 2)^2/[(x - 2)(x + 2)] = (x - 2)/(x + 2)`
b, Để `A = 1/2 <=> (x - 2)/(x + 2) = 1/2`
`<=> 2(x - 2) = x + 2`
`<=> 2x - 4 = x + 2`
`<=> 2x - x = 2 + 4`
`<=> x= 6 (TM)`
2. Ta có
`P = (x^2 + 2x)/(2x + 10) + (x - 5)/x + (50 - 5x)/[2x(x + 5)]`
`= (x^2 + 2x)/[2(x + 5)] + (x- 5)/x + (50 - 5x)/[2x(x + 5)]`
`= [x(x^2 + 2x)]/[2x(x + 5)] + [2(x - 5)(x + 5)]/[2x(x + 5)] + (50 - 5x)/[2x(x + 5)]`
`= (x^3 + 2x^2 + 2x^2 - 50 + 50 - 5x)/[2x(x + 5)]`
`= (x^3 + 4x^2 - 5x)/[2x(x + 5)]`
`= [(x - 1)x(x + 5)]/[2x(x + 5)]`
`= (x - 1)/2`
Bài 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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