Question

$determinati \: vaorile \: intregi \: ale \: lui \: n \: pentru \: care \: \frac{4n + 23}{2n + 1} \: aprtine \: multimii \: z$
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Answer 1

Răspuns:

Explicație pas cu pas:

(4n+23)/(2n+1) ∈ Z ⇔ (4n+2+21)/(2n+1) = [2(2n+1) + 21] / (2n+1) = 2 + 21/(2n+1)

⇒21/(2n+1) ∈ Z ⇔ 2n+1 ∈ {-21,-7,-3,-1,1,3,7,21}

2n+1 = -21 ⇒ 2n=-22, n=-11 ∈Z

2n+1 = -7 ⇒ 2n=-8, n=-4 ∈ Z

2n+1 = -3 ⇒ 2n=-4, n=-2 ∈ Z

2n+1 = -1 ⇒ 2n=-2, n=-1 ∈ Z

2n+1= 1 ⇒ 2n = 0, n=0 ∈ Z

2n+1=3 ⇒ 2n=2, n=1 ∈ Z

2n+1=7 ⇒2n=6, n=3∈Z

2n+1=21 ⇒2n=20, n=10∈Z

deci n∈{-11,-4,-2,-1,0,1,3,10}