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a: Để A nguyên thì \(2n+1\inƯ\left(10\right)\)
mà n nguyên
nên \(2n+1\in\left\{1;-1;5;-5\right\}\)
=>\(n\in\left\{0;-1;2;-3\right\}\)
b: B nguyên thì 3n+5-5 chia hết cho 3n+5
=>\(3n+5\inƯ\left(-5\right)\)
mà n nguyên
nên \(3n+5\in\left\{-1;5\right\}\)
=>n=-2 hoặc n=0
c: Để C nguyên thì 4n-6+16 chia hết cho 2n-3
=>\(2n-3\in\left\{1;-1\right\}\)
=>\(n\in\left\{2;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{8-3n}{5-3n}\inℤ\)
\(\Rightarrow\frac{3+5-3n}{5-3n}\inℤ\)
\(\Rightarrow\frac{3}{5-3n}+\frac{5-3n}{5-3n}\inℤ\)
\(\Rightarrow\frac{3}{5-3n}+1\inℤ\Leftrightarrow\frac{3}{5-3n}\inℤ\)
\(\Rightarrow3⋮5-3n\)
\(\Rightarrow5-3n\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow3n\in\left\{\pm6;\pm8\right\}\)
\(\Rightarrow\hept{\begin{cases}n=6:3\\n=8:3\left(\notinℤ\right)\end{cases}}\Leftrightarrow\hept{\begin{cases}n=2\\n=\frac{8}{3}\left(loại\right)\end{cases}}\)
\(\Rightarrow n=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{10-3n}{5-3n}\inℤ\Leftrightarrow\frac{10-n}{5-n}\inℤ\)
\(\Rightarrow\frac{5-n}{n}\inℤ\Leftrightarrow n\in\left\{1;-1;5;-5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Vì -3; n- 1 nên M là phân số nếu n – 1 khác 0 => n khác 1
b) Với n = 3 => M = − 3 3 − 1 = − 3 2
Với n = 5 => M = − 3 5 − 1 = − 3 4 và n = -4 => M = − 3 − 4 − 1 = − 3 − 5
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3n-5}{4n+1}\)ϵ z =>\(\dfrac{4\left(3n-5\right)}{4n+1}\)ϵ z
Ta có :
\(\dfrac{4\left(3n-5\right)}{4n+1}\)=\(\dfrac{3\left(4n+1\right)-23}{4n+1}\)=3-\(\dfrac{23}{4n+1}\)
Để \(\dfrac{4\left(3n-5\right)}{4n+1}\)ϵ Z=>4n+1ϵ Ư(23)=(1;-1;23;-23)
4n+1=1=>n=0
4n+1=-1=>n=\(\dfrac{-1}{2}\)(loại)
4n+1=23=>n=\(\dfrac{11}{2}\)(loại)
4n+1=-23=>n=-6
Vậy n ϵ 0;-6
\(\dfrac{3n-5}{4n+1}\) là số nguyên khi :
\(3n-5⋮4n+1\)
\(\Rightarrow4\left(3n-5\right)-3\left(4n+1\right)⋮4n+1\)
\(\Rightarrow12n-20-12n-3⋮4n+1\)
\(\Rightarrow-23⋮4n+1\)
\(\Rightarrow4n+1\in\left\{-1;1;-23;23\right\}\)
\(\Rightarrow n\in\left\{\dfrac{1}{2};0;-6;\dfrac{11}{2}\right\}\Rightarrow n\in\left\{0;-6\right\}\left(n\in Z\right)\)
M là số nguyên `<=> 10-3n \vdots 5-3n`
`<=> (5-3n)+5 \vdots (5-3n)`
`<=> 5 \vdots (5-3n)`
`<=> (5-3n) \in Ư(5)`
`<=> 5-3n \in {-5;5;-1;1}`
`<=> -3n \in {-10;0;-5;-4}`
`<=> n \in {10/3 ; 0 ; 5/3 ; 4/3}`
Để M là số nguyên thì 10-3n⋮5-3n
5+5-3n⋮5-3n
5-3n⋮5-3n
⇒5⋮5-3n ⇒5-3n∈Ư(5)
Ư(5)={-1;1;-5;5}
⇒n∈{2;0}