`@` `\text {Ans}`
`\downarrow`
`b)`
`x - 37 + 63 = -33`
`=> x - (37 - 63) = -33`
`=> x - (-26) = -33`
`=> x+ 26 = -33`
`=> x = -33 - 26`
`=> x =-59`
Vậy, `x = -59`
`c)`
`30 + 5(-9 - x) = -35`
`=> 5(-9-x) = -35 - 30`
`=> 5(-9-x) = -65`
`=> -9 - x = -65 \div 5`
`=> - 9 - x = -13`
`=> x = - 9 - (-13)`
`=> x = - 9 + 13`
`=> x = 4`
Vậy, `x = 4`
`d)`
`(x + 1)(4 - x) > 0`
`=> `\(\left[{}\begin{matrix}\left(x+1\right)>0\\\left(4-x\right)>0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\left(x+1\right)-1>0-1\\\left(4-x\right)>0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x>-1\\x< 4\end{matrix}\right.\)
Hay `-1 < x < 4`
`e)`
`(3 - x)(y + 2) = 5`
`=> (3 - x)(y + 2) \in` Ư`(5) = {1; 5; -1; -5}`
Ta có bảng sau:
| `3 - x` | `1` | `5` | `-1` | `-5` |
| `y + 2` | `5` | `1` | `-5` | `-1` |
| `x` | `2` | `-2` | `4` | `8` |
| `y` | `3` | `-1` | `-7` | `-3` |
Vậy, ta có cặp `(x; y)` thỏa mãn \(\left\{2;3\right\};\left\{-2;-1\right\};\left\{-7;4\right\};\left\{-3;8\right\}.\)
