Ta có: \(\overline{ab}=\overline{ba}.4+a\)
\(\Rightarrow10a+b=4\left(10b+a\right)+a\)
\(\Leftrightarrow10a+b=40b+5a\)
\(\Leftrightarrow5a-39b=0\)\(\Leftrightarrow a=\frac{39b}{5}\)
Vì a,b\(\in N,10>a>0,10>b\ge0\) nên
\(39b⋮5\)
\(\Rightarrow b⋮5\Rightarrow b=5\Rightarrow a=39\left(KTM\right)\)
Vậy ko tồn tại a,b TM