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