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\(a.\dfrac{12}{3}=\dfrac{20}{5}=4\\ b.\dfrac{9}{-3}=\dfrac{-15}{5}=-3\)
a, Xét \(\dfrac{x}{3}=4\Rightarrow x=12;\dfrac{20}{y}=4\Rightarrow y=\dfrac{20}{4}=5\)
b, \(\dfrac{9}{-x}=-3\Rightarrow-x=-3\Leftrightarrow x=3\)
\(\dfrac{y}{5}=-3\Rightarrow y=-15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(xy-2x+y=1\)
\(\Leftrightarrow x\left(y-2\right)+\left(y-2\right)=-1\)
\(\Leftrightarrow\left(x+1\right)\left(y-2\right)=-1\)
Ta có bảng sau:
\(x+1\) | 1 | -1 |
\(y-2\) | -1 | 1 |
\(x\) | 0 | -2 |
\(y\) | 1 | 3 |
Vậy ta tìm được các cặp số \(\left(0;1\right);\left(-2;3\right)\) thỏa yêu cầu bài toán.
![](https://rs.olm.vn/images/avt/0.png?1311)
x2.(y+1) + y = 30
x2. (y+1) + (y+1) = 29
(y+1).(x2+1) = 29 = 1 . 29 = 29 . 1
![](https://rs.olm.vn/images/avt/0.png?1311)
x-3/-9 = -4/x-3
(x-3)^2=-9.-4=36
=>TH1 x-3= 6
TH2 x-3 = -6
=> TH1: X= 9
TH2: X=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(2y\left(x+1\right)-x-7=0\)
\(\Rightarrow2y\left(x+1\right)-x-1-6=0\)
\(\Rightarrow2y\left(x+1\right)-\left(x+1\right)=6\)
\(\Rightarrow\left(x+1\right)\left(2y-1\right)=6\)
..........
Chia ra các trường hợp em nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)5/8x+2/5=1/5`
`=>5/8x=1/5-2/5`
`=>5/8x=-1/5`
`=>x=-1/5:5/8=-8/25`
`b)5/7:x+11/7=18/7`
`=>5/7:x=18/7-11/7`
`=>5/7:x=1`
`=>x=5/7`
`c)(-1,2).(-3/24)+(0,4-1 4/15):1 2/3`
`=(-6/5).(-1/8)+(2/5-19/15):5/3`
`=3/20+(-13/15)*3/5`
`=3/20-13/25=-37/100`
a)5/8.x+2/5=1/5
5/8.x=1/5 - 2/5
5/8.x=-1/5
x=(-1/5):5/8
x=(-1/5).8/5
x=-8/25. Vậy x=-8/25
b)5/7:x +11/7=18/7
5/7:x=1
x=5/7:1
x=5/7. Vậy x=5/7
x,y∈Z⇒x+1,xy-1∈Z và x+1,xy-1 thuộc Ư(3)
Ta có bảng:
Vậy \(\left(x,y\right)\in\left\{\left(2;1\right)\left(-2;1\right);\left(-4;0\right)\right\}\)
Vì x,y là chữ số nên \(x,y\in\left\{0;1;2;3;4;5;6;7;8;9\right\}\)
Ta có \(\left(x+1\right)\left(xy-1\right)=3=3.1=1.3\)
Với \(\left\{{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\-1=3\end{matrix}\right.\left(loại\right)\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)