Ta có:
\(x+3⋮x-1\)
\(\Rightarrow\left(x-1\right)+4⋮x-1\)
\(\Rightarrow4⋮x-1\)
\(\Rightarrow x-1\in U\left(4\right)=\left\{-1;1;-2;2;-4;4\right\}\)
\(\Rightarrow x\in\left\{0;2;-1;3;-3;5\right\}\)
Vậy \(x\in\left\{0;2;-1;3;-3;5\right\}\)
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\(x+3⋮x-1.\)
\(\left(x-1\right)+4⋮x-1.\)
mà \(x-1⋮x-1.\)
\(\Rightarrow4⋮x-1.\)
\(\Rightarrow x-1\inƯ_{\left(4\right)}=\left\{\pm1;\pm2;\pm4\right\}.\)
Ta có bảng giá trị sau:
| x - 1 | -4 | -2 | -1 | 1 | 2 | 4 |
| x | -3 | -1 | 0 | 2 | 3 | 5 |
Vậy \(x\in\left\{-3;-1;0;2;3;5\right\}.\)
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