![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: \(2x-1=2\left(x-3\right)+5\)
để \(2x-1⋮x-3\Rightarrow2\left(x-3\right)+5⋮x-3\\ m\text{à }x.nguy\text{ê}n\Rightarrow x-3nguy\text{ê}n\\ \Rightarrow x-3\in\text{Ư}\left(5\right)=\left\{-5;5;1;-1\right\}\)
ta có bảng sau :
x-3 | -5 | 5 | -1 | 1 |
x | -2 | 2 | 4 | 8 |
\(\Leftrightarrow2.\left(x-3\right)+5⋮x-3\)
\(do2.\left(x-3\right)⋮x-3\)
\(\Leftrightarrow5⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, n+5 chia hết cho n+2
n+2 chia hết cho n+2
=> (n+5) - (n+2) chia hết cho 2
n+5-n-2 chia hết cho 2
3 chia hết cho 2
=>2 thuộc Ư(3)=...
b, 2n+1 chia hết cho n+5
n+5 chia hết cho n+5 => 2(n+5) chia hết cho n+5
Làm tương tự ý a
c, n2+3n+13 = n (n+3) +13
Mà n(n+3) chia hết cho n+3
=> 13 chia hết cho n+3
=> n+3 thuộc Ư(13)
=>...
![](https://rs.olm.vn/images/avt/0.png?1311)
<=>\(\left|\dfrac{1}{2}-2x\right|=\dfrac{5}{3}< =>\left[{}\begin{matrix}\dfrac{1}{2}-2x=\dfrac{5}{3}\\\dfrac{1}{2}-2x=\dfrac{-5}{3}\end{matrix}\right.< =>\left[{}\begin{matrix}2x=\dfrac{-7}{6}\\2x=\dfrac{13}{6}\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{-7}{12}\\x=\dfrac{13}{12}\end{matrix}\right.\)
\(\left|\dfrac{1}{2}+2x\right|+\dfrac{2}{3}=\dfrac{7}{3}\)
\(\left|\dfrac{1}{2}+2x\right|=\dfrac{7}{3}-\dfrac{2}{3}=\dfrac{5}{3}\)
⇔\(\left[{}\begin{matrix}\dfrac{1}{2}+2x=\dfrac{5}{3}\\\dfrac{1}{2}+2x=-\dfrac{5}{3}\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow n-3\in\left\{-1;1;11\right\}\)
hay \(n\in\left\{2;4;14\right\}\)
Lời giải:
a.
$3n+2\vdots n-3$
$3(n-3)+11\vdots n-3$
$\Rightarrow 11\vdots n-3$
$\Rightarrow n-3\in\left\{1; -1; 11; -11\right\}$
$\Rightarrow n\in\left\{4; 2; 14; -8\right\}$
Vì $n$ tự nhiên nên $n\in\left\{4;2;14\right\}$
b.
$n^2+7n+9\vdots n+7$
$n(n+7)+9\vdots n+7$
$\Rightarrow 9\vdots n+7$
$\Rightarrow n+7\in\left\{1; -1; 3; -3; 9; -9\right\}$
$\Rightarrow n\in\left\{-6; -8; -4; -10; 2; -16\right\}$
Vì $n$ tự nhiên nên $n=2$
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow n-3\in\left\{-1;1;11\right\}\)
hay \(n\in\left\{2;4;14\right\}\)
Sửa đề: Tìm n để A nguyên
Để A nguyên thì n-3+5 chia hết cho n-3
=>n-3 thuộc {1;-1;5;-5}
=>n thuộc {4;2;8;-2}