Ta có:
ab x aba = abab
ab x aba = ab x100 + ab
ab x aba = ab x (100 + 1)
ab x aba = ab x 101
=> aba = 101
=> a = 1 ; b = 0
Ta có : ab x aba = abab
ab x aba = ab x 100 x ab
ab x aba = ab x ( 100 + 1)
ab x aba = ab x 101
=> aba = 101
=> a = 1 ; b = 0
ab.aba=abab⇔ababa=abab⇔ababa−abab=0
⇔abab(a−1)=0⇔{abab=0a−1=0 ⇔{a=1bb=0 ⇔{a=1b=0
vậy a=1;b=0
ab x aba = abab
ab x aba = ab x 100 + ab
ab x aba = ab x ( 100 + 1 )
ab x aba = ab x 101
=> ab = 101
Vậy a = 1 , b = 0
Ta có : ab x aba = abab
ab x aba = ab x 101
Do đó : aba = 101
Vậy a = 1 ; b = 0
\(ab\cdot aba=abab\)
\(\Rightarrow ab\cdot aba=ab\cdot100+ab\)
\(\Rightarrow ab\cdot aba=ab\cdot(100+1)\)
\(\Rightarrow ab\cdot aba=ab\cdot101\)
Cả hai vế đều bỏ bớt ab . Nên aba = 101
=> a = 1 , b = 0