![](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)
Lời giải:
Đặt $3x+5y=a; x+4y=b$.
Ta có: $2a+b=2(3x+5y)+x+4y=7x+14y=7(x+2y)\vdots 7$
$ab\vdots 7\Rightarrow a\vdots 7$ hoặc $b\vdots 7$
Nếu $a\vdots 7$. Khi đó: $2a+b\vdots 7\Rightarrow b\vdots 7$
$\Rightarrow ab\vdots (7.7)$ hay $ab\vdots 49$
Nếu $b\vdots 7$. Khi đó: $2a+b\vdots 7\Rightarrow 2a\vdots 7\Rightarrow a\vdots 7$
$\Rightarrow ab\vdots (7.7)$ hay $ab\vdots 49$
Vậy ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Thêm cả 100 vào nữa nha bạn.
\(n\in\left\{1;2;4;5;10;20;25;50;100\right\}\)
Để x là số nguyên khi:
\(100⋮x\)
Hay \(x\inƯ\left(100\right)=\left\{1;\pm2;\pm4;\pm5;\pm10;\pm20;\pm25;\pm50;\pm100\right\}\)
Vậy x thuộc các số trên thì x là số nguyên.
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(3x+2xy=7\)
\(\Leftrightarrow x\left(2y+3\right)=7\)
\(\Leftrightarrow x;2y+3\inƯ\left(7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\2y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\2y+3=-7\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\2y+3=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\2y+3=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\y=-\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
b/ \(3x-5xy=11\)
\(\Leftrightarrow x\left(3-5y\right)\inƯ\left(11\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\3-5y=11\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\3-5y=-11\end{matrix}\right.\\\left\{{}\begin{matrix}x=11\\3-5y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-11\\3-5y=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-\dfrac{8}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=\dfrac{14}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=7\\y=\dfrac{2}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-7\\y=-\dfrac{4}{5}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
a) A= (1;2;3;4;5)
b) B= ( 63;64;65;66;67;68;69;70)
Bài 2 :
a) 10x-5 = 11.5-10
10x-5 = 55-10
10x=45+5
10x=50
x=5
b) 27-3x=9.2-3
27-3x = 18-3
27-3x=15
3x=27-15
3x=12
x=4
c) 4x-15=12:12
4x-15=1
4x=16
x=4
d) 2+13x=14.2
13x=28-2
13x=26
x=2
a) \(10x-5=45\)
\(10x=40\)
\(x=4\)
b) \(27-3x=15\)
\(3x=27-15=12\)
\(x=\dfrac{12}{3}=4\)
c) \(4x-15=1\)
\(4x=16\)
\(x=\dfrac{16}{4}=4\)
d) \(2+13x=28\)
\(13x=26\)
\(x=\dfrac{26}{13}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình làm câu khó thôi nhé.
2x chia hết cho 3
=>(2x+x-x) chia hết cho 3
=>(3x-x) chia hết cho 3
3x chia hết cho 3=>x chia hết cho 3
=>x thuộc B(3)={0;3;6;...}
Vậy x thuộc {0;3;6;...ư}
3x+2ϵƯ(60)
=>3x+2ϵ{-1;-2;-3;-4;-5;-6;-10;-12;-15;-20;-30;-60;
1;2;3;4;5;6;10;12;15;20;30;60}
Ta thấy:3x chia 3 dư 0
2 chia 3 dư 2
=>3x+2 chia 3 dư 2
=>3x+2ϵ{............}
=>3xϵ{..........}
=>xϵ{..........}
Vậy.........