\(36⋮6x\) => \(6x\in U\left(36\right)=\left\{1;2;3;4;6;9;12;18;36\right\}\)
Xét TH1: 6x = 1 => \(x=\frac{1}{6}\)
Xét TH2: 6x = 2 => \(x=\frac{1}{3}\)
Xét TH3: 6x = 3 => \(x=\frac{1}{2}\)
Xét TH4: 6x = 4 => \(x=\frac{2}{3}\)
Xét TH5: 6x = 6 => x = 1
Xét TH6: 6x = 9 => \(x=\frac{3}{2}\)
Xét TH7: 6x = 12 => x = 2
Xét TH8: 6x = 18 => x = 3
Xét TH9: 6x = 36 => x = 6
Vậy \(x\in\left\{\frac{1}{6};\frac{1}{3};\frac{1}{2};\frac{2}{3};1;\frac{3}{2};2;3;6\right\}\)
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