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Cho A =x^2+16/x+3 tìm x nguyên để A nguyên . Mọi người giúp em vs ạ ! Em cảm ơn ạ !
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Đáp án+Giải thích các bước giải:
`A in Z`
`->x^2+16 vdots x+3`
`->x^2-9+25 vdots x+3`
`->(x-3)(x+3)+25 vdots x+3`
`->25 vdots x+3`
`->x+3 in Ư(25)={+-1,+-5,+-25}`
`->x in {-2,-4,2,-8,22,-28}`
`A = (x^2 +16)/(x+3) = (x^2 – 9 + 25)/(x+3) = ((x-3)(x+3) – 25)/(x+3) = ((x-3)(x+3))/(x+3) – 25/(x+3) = x – 3 – 25/(x+3) ĐKXĐ : x` $\neq$ `-3`
Vì `x ∈ Z` `⇒ x – 3 ∈ Z` `⇒ A ∈ Z` khi:
`25/(x+3) ∈ Z`
`⇒ 25 vdots x + 3`
`⇔ x + 3 ∈{±1;±5±25}`
`⇒ x ∈{22,2,-2,-4,-8,-28}`