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Có a,b = b,a x 3 + 1,3
=> ab = ba x 3 + 1,3
=> 10a + b = 30b + 3a + 13
=> 7a - 29b = 13
=> 7a = 29b + 13
Mà \(b>0=>29b+13>13\)
=> 7a > 13
=> a \(\ge2\)
TH1: a = 2 => b = \(\dfrac{1}{29}\left(l\right)\)
TH2: a = 3 => b = \(\dfrac{8}{29}\left(l\right)\)
TH3: a = 4 => b = \(\dfrac{15}{29}\left(l\right)\)
TH4: a = 5 => \(b=\dfrac{22}{29}\) (l)
TH5: a = 6 => \(b=1\) (thỏa mãn)
TH6: a = 7 => \(b=\dfrac{36}{29}\left(l\right)\)
TH7: a = 8 => b = \(\dfrac{43}{29}\left(l\right)\)
TH8: a = 9 => \(b=\dfrac{50}{29}\left(l\right)\)
Vậy a = 6, b = 1