Tìm số nguyên n sao cho 10n + 13 chia hết cho n + 1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(10n+13=10\left(n+1\right)+3\)
Để 10n+13 chia hết cho n+1 thì 10(n+1)+3 chia hết cho n+1
=> 3 chia hết cho n+1 (1)
Vì \(n\inℕ\Rightarrow n+1\inℕ\)(2)
(1)(2) => n+1\(\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Ta có bảng giá trị
n+1 | -3 | -1 | 1 | 3 |
n | -4 | -2 | 0 | 2 |
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n^2+3⋮n-1\)
\(\Rightarrow n\left(n-1\right)+n+3⋮n-1\)
\(\Rightarrow n+3⋮n-1\)
\(\Rightarrow\left(n-1\right)+4⋮n-1\)
\(\Rightarrow4⋮n-1\)
\(\Rightarrow n-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{2;0;3;-1;5;-3\right\}\)
Vậy.......................................
![](https://rs.olm.vn/images/avt/0.png?1311)
a) n^2 + 3n - 13 chia hết cho n + 3
n(n + 3) - 13 chia hết cho n + 3
n(n + 3) chia hết cho n + 3
Nên 13 chia hết cho n + 3
Tự tìm nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>\(n+2\in\left\{1;-1;7;-7\right\}\)
=>\(n\in\left\{-1;-3;5;-9\right\}\)
b: =>n-3+4 chia hết cho n-3
=>\(n-3\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(n\in\left\{4;2;5;1;7;-1\right\}\)
c: =>3n^3+n^2+9n^2-1-4 chia hết cho 3n+1
=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)
d: =>10n^2-10n+11n-11+1 chia hết cho n-1
=>\(n-1\in\left\{1;-1\right\}\)
=>\(n\in\left\{2;0\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Answer:
a) \(\left(n+2\right)⋮\left(n-3\right)\)
\(\Rightarrow\left(n-3+5\right)⋮\left(n-3\right)\)
\(\Rightarrow5⋮\left(n-3\right)\)
\(\Rightarrow n-3\) là ước của \(5\), ta có:
Trường hợp 1: \(n-3=-1\Rightarrow n=2\)
Trường hợp 2: \(n-3=1\Rightarrow n=4\)
Trường hợp 3: \(n-3=5\Rightarrow n=8\)
Trường hợp 4: \(n-3=-5\Rightarrow n=-2\)
b) Ta có: \(x-3\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
\(\Rightarrow x\in\left\{4;16;2;-10\right\}\)
Vậy để \(x-3\inƯ\left(13\right)\Rightarrow x\in\left\{4;16;2;-10\right\}\)
c) Ta có: \(x-2\inƯ\left(111\right)\)
\(\Rightarrow x-2\in\left\{\pm111;\pm37;\pm3;\pm1\right\}\)
\(\Rightarrow x\in\left\{-99;-35;1;1;3;5;39;113\right\}\)
d) \(5⋮n+15\Rightarrow n+15\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Trường hợp 1: \(n+15=-1\Rightarrow n=-16\)
Trường hợp 2: \(n+15=1\Rightarrow n=-14\)
Trường hợp 3: \(n+15=5\Rightarrow n=-10\)
Trường hợp 4: \(n+15=-5\Rightarrow n=-20\)
Vậy \(n\in\left\{-14;-16;-10;-20\right\}\)
e) \(3⋮n+24\)
\(\Rightarrow n+24\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow n\in\left\{-23;-25;-21;-27\right\}\)
f) Ta có: \(x-2⋮x-2\)
\(\Rightarrow4\left(x-2\right)⋮x-2\)
\(\Rightarrow4x-8⋮x-2\)
\(\Rightarrow\left(4x+3\right)-\left(4x-8\right)⋮x-2\)
\(\Rightarrow11⋮x-2\)
\(\Rightarrow x-2\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
\(\Rightarrow x\in\left\{3;13;1;-9\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt \(A=\frac{3n-13}{n+3}=\frac{3\left(n+3\right)-22}{n+3}=3-\frac{22}{n+3}\)
=> 22 \(⋮\)n + 3 => n + 3 \(\in\)Ư(22) = { \(\pm1;\pm2;\pm11;\pm22\)}
n + 3 | 1 | -1 | 2 | -2 | 11 | -11 | 22 | -22 |
n | -2 | -4 | -1 | -5 | 8 | -14 | 19 | -25 |
b) Đặt \(B=\frac{2n+3}{n-1}=\frac{2\left(n-1\right)+5}{n-1}=2+\frac{5}{n-1}\)
=> 5 \(⋮\)n - 1 => n - 1 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
n - 1 | 1 | -1 | 5 | -5 |
n | 2 | 0 | 6 | -4 |
\(\left(a\right)3n-13⋮n+3\)
\(3n-13=3\left(n+3\right)-22\)
\(=>n+3=Ư\left(22\right)\)
\(n+3=\left\{-22;-11;-2;-1;1;2;11;22\right\}\)
\(=>n=\left\{-25;-14;-5;-4;-2;-1;8;19\right\}\)
\(\left(b\right)2n+3⋮n-1\)
\(2n+3=2\left(n-1\right)+5\)
\(=>n-1=Ư\left(5\right)\)
\(n-1=\left\{-5;-1;1;5\right\}\)
\(=>n=\left\{-4;0;2;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2n+1}{n-5}=\frac{2n-10+11}{n-5}=\frac{2n-10}{n-5}+\frac{11}{n-5}=2+\frac{11}{n-5}\)
=> 11 chia hết cho n-5
n-5 thuộc Ư (11) = { -11; -1; 1; 11}
( rồi bạn thế vô rồi tính nha ^^ ... tương tự đối với b và c)