a) Ta có : ab + ba = a . 10 + b + b . 10 + a
= a . (10 + 1) + b . ( 1 + 10)
= a . 11 + b . 11
= (a + b) . 11 \(⋮\)11
b)Ta có : abc - cba = (a . 100 + b . 10 + c) - (c . 100 + b . 10 + a)
= a . 100 + b . 10 + c - c . 100 - b . 10 - a
= a . (100 - 1) + (b . 10 - b . 10) + c . (1 - 100)
= a . 99 + 0 + c . ( - 99)
= (a - c) . 99 \(⋮\)99
c) tự làm
a) ab + ba = 10a + b + 10b + a
= 11a + 11b
= 11(a + b) \(⋮\)11
=> ab + ba \(⋮\)11.
b)abc - cba = 100a + 10b + c - 100c - 10b - a
= 99a - 99c
= 99(a - c) \(⋮\)99
=> abc - cba \(⋮\)99
c)aaa + bbb = 100a + 10a + a + 100b + 10b + b
= 100(a +b) + 10(a + b) + (a + b)
= (a + b)(100 + 10 + 1)
= (a + b) 111
= (a + b) . 3 . 37 \(⋮\)37
=> aaa + bbb \(⋮\)37
a, ab+ba= a0+b+b0+a=a(10+1)+b(10+1)=11a+11b=11(a+b)
b, abc-cba= a00+b0+c-c00-b0-a=a(100-1)+b(10-10)--(100-1)c=99a-99c=99(a-c)
c, aaa+bbb= a00+a0+a+b00+b0+b=(a+b)100+(a+b)10+(a+b)1=(a+b)(100+10+1)=111(a+b)=37.3(a+b)
a) ab + ba = a.10 + b + b.10 + a = (a.10 + a) + ( b.10 + b) = a.(10+1) + b.(10+1) = a.11+b.11 \(=\left(a+b\right).11⋮11\)(ĐPCM)
b) abc - cba = a.100 + b.10 + c - c.100 - b.10 - a = (a.100 - a) + (b.10 - b.10) + (c - c.100) = a.(100-1) + 0 + c.(1 - 100) = a.99 - c.99 \(=\left(a-c\right).99⋮99\)(ĐPCM)
c) aaa + bbb = a.111+b.111= \(=\left(a+b\right).111=\left(a+b\right).3.37⋮37\)(ĐPCM)
