ab = 10 (a + b)
ba = 10 (b + a)
=> ab + ba = 11 (a+b) chia hết cho 11
\(\overline{ab}+\overline{ba}\)
\(=10a+b+10b+a\)
\(=\left(10+1\right)a+\left(10+1\right)b\)
\(=11a+11b\)
\(=11\left(a+b\right)⋮11\) (đpcm)
ab+ba chia hết cho 11
ab=10a+b ba=10b+a
ab+ba=10a+b+10b+a
=(10a+a)+(10b+b)
=11a+11b
vậy:ab+ba chia hết cho 11(vì mik chứng minh a chia hết cho b khi a=k.b)
\(\text{Ta có : }\\ \dfrac{ }{ab}_+\dfrac{ }{ba}=\left(10a+b\right)+\left(10b+a\right)\\ =10a+b+10b+a\\ =\left(10a+a\right)+\left(b+10b\right)\\ =11a+11b\\ =11\left(a+b\right)\text{ }⋮\text{ }11\\ \text{ }\Rightarrow\dfrac{ }{ab}_+\dfrac{ }{ba}\text{ }⋮\text{ }11\text{ }\left(ĐPCM\right)\\ \text{ Vậy }\dfrac{ }{ab}_+\dfrac{ }{ba}\text{ }⋮\text{ }11\)
