Ta có : \(\overline{ab}+\overline{ba}=132\)
\(\Rightarrow10a+b+10b+a=132\)
\(\Rightarrow11a+11b=132\)
\(\Rightarrow11\left(a+b\right)=132\)
\(\Rightarrow a+b=\dfrac{132}{11}=12\) mà có a - b = 4
\(\Rightarrow a=\left(12+4\right):2=8\)
\(\Rightarrow b=8-4=4\)
\(\Rightarrow\overline{ab}=84\)
Ta có : \(\overline{ab}+\overline{ba}=132\)
<=> 10a + b + 10b + a = 132
<=> 11a + 11b = 132
<=> 11(a + b) = 132
<=> a + b = 12 (1)
Lại có a - b = 4 (2)
Từ (1),(2) => a = (12 + 4) : 2 = 8
b = 12 - 8 = 4
a,b thỏa mãn
@Học 24
\(\overline{ab}+\overline{ba}=132\)
\(\Rightarrow10a+b+10b+a=132\)
\(\Rightarrow11a+11b=132\)
\(\Rightarrow11\left(a+b\right)=132\)
\(\Rightarrow a+b=12\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=12\\a-b=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b+a-b=16\Leftrightarrow2a=16\Leftrightarrow a=8\\b=8-4=4\end{matrix}\right.\)
