2) Tìm x :
a) \(3x-7⋮x+2\)
Ta có : \(x+2⋮x+2\Rightarrow3x+6⋮x+2\)
\(\Rightarrow3x-7-\left(3x+6\right)⋮x+2\)
\(\Rightarrow-13⋮x+2\) hay \(x+2\inƯ\left(-13\right)=\left\{1,-1,13,-13\right\}\)
\(\Rightarrow x\in\left\{-1,-3,11,-15\right\}\)
Vậy : \(x\in\left\{-1,-3,11,-15\right\}\)
b) \(\left(4x+5\right)⋮\left(x-11\right)\)
Ta có : \(x-11⋮x-11\Rightarrow4x-44⋮x-11\)
\(\Leftrightarrow4x+5-\left(4x-44\right)⋮x-11\)
\(\Leftrightarrow49⋮x-11\) hay \(x-11\inƯ\left(49\right)=\left\{1,-1,49,-49\right\}\)
\(\Leftrightarrow x\in\left\{12,10,60,-38\right\}\)
Vậy : \(x\in\left\{12,10,60,-38\right\}\)
c) \(xy+2x-y=4\)
\(\Leftrightarrow x\left(y+2\right)-\left(y+2\right)=4-2=2\)
\(\Leftrightarrow\left(x-1\right).\left(y+2\right)=2\)
Do \(x\in Z\Rightarrow\left\{{}\begin{matrix}x-1\in Z\\y+2\in Z\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)\) và \(\left(y+2\right)\) là các cặp ước của 2
Ta có bảng giá trị sau :
| \(x-1\) | 2 | -2 | 1 | -1 |
| \(x\) | 3 | -1 | 2 | 0 |
| \(y+2\) | 1 | -1 | 2 | -2 |
| \(y\) | -1 | -3 | 0 | -4 |
| Đánh giá | Chọn | Chọn | Chọn | Chọn |
Vậy : \(\left(x,y\right)\in\left\{\left(3,-1\right);\left(-1,-3\right);\left(2,0\right);\left(0,-4\right)\right\}\)