\(ab.aba=abab\)
\(\Rightarrow10a+b+101a+10b=1010a+101b\)
\(\Leftrightarrow111a+11b=1010a+101b\)
ab x aba =abab
aba=ab x101:ab
aba=101
a=1
b=0
ab x aba = abab
aba=ab x 101 : ab
aba=101
=> a=1 b=0
ta có :
ab x aba = abab
ab x aba = ab x 100 + ab
ab x aba = ab x ( 100 + 1 )
ab x aba = ab x 101
\(\rightarrow\)aba = 101
vậy a = 1 ; b = 0