​Ta có :

\(\overline{aaa}+\overline{bbb}=100a+10a+a+100b+10b+b=111a+111b=111\left(a+b\right)⋮11\left(dpcm\right)\)

\(\overline{aaa}+\overline{bbb}⋮37\)

\(\Rightarrow a.111+b.111⋮37\)

mà \(111⋮37\)

\(\Rightarrow\) 111.( a+b ) \(⋮\) 37

\(\Rightarrow\overline{aaa}+\overline{bbb}⋮37\)

\(\overline{aaa}+\overline{bbb}=111a+111b=111\left(a+b\right)=37.3\left(a+b\right)\)

Vậy \(\overline{aaa}+\overline{bbb}⋮37\)

Chúc hok dốt

\(\overline{aaa}+\overline{bbb}=111.a+111.b=111\left(a+b\right)=37\times3\times\left(a+b\right)⋮37\)

(aaa + bbb) = 111a + 111b = 111( a + b )
Vì 111 chia hết cho 37 => ( a + b ) chia hết cho 37
=> ( aaa + bbb ) chia hết cho 37

(aaa+bbb):37

(a x 100 + a x 10 + a + b x 100 + b x 10 + b ):37

(a x (100 + 10 +1 ) + b x (100 + 10 + 1 ) : 37

(a x 111 + b x 111):37

(111 x (a + b) :37

( 37 x 3 x (a + b) :37 

vậy aaa + bbb : 37

aaa + bbb=(100a+b)+(100b+a)

              = 111a+111b

              = 111.(a+b)

nếu đúng các bạn nhớ cho mình nhé

Ta có:

\(\overline{abcabc}=1001\overline{abc}\)

\(=143.7.\overline{abc}\)

\(\Rightarrow1001\overline{abc}⋮7\Rightarrow\overline{abcabc}⋮7\)

\(\rightarrowđpcm\)

\(\overline{aaa}=111a\)

\(=37.3.a\)

\(\Rightarrow111a⋮37\Rightarrow\overline{aaa}⋮37\)

\(\rightarrowđpcm\)

\(\overline{1ab1}-\overline{1ba1}\)

\(=1000+\overline{ab}+1-1000-\overline{ba}-1\)

\(=\overline{ab}-\overline{ba}\)

\(=10a+b-10b-a\)

\(=9a-9b\)

\(=9\left(a-b\right)⋮9\)

Mà \(\overline{1ab1}-\overline{1ba1}=\overline{...0}⋮10\)

\(\Rightarrow\overline{1ab1}-\overline{1ba1}⋮9;10\Rightarrow⋮90\)

\(\rightarrowđpcm\)

b, ta có \(\overline{aaa}\)=111.a=37.3.a \(⋮\)37

=> aaa chia hết 37 (đpcm)

Ta có \(\overline{aaa}=a.111=a.3.37\)

\(=>a.3.37⋮37\)

Vậy \(\overline{aaa}⋮37\left(dpcm\right)\)

Ta có

Vậy

đề a,b bạn viết sai

c,\(\overline{abcabc}\) :7

Theo bài ra, ta có:

\(\overline{abcabc}\) = 1000\(\overline{abc}\) + \(\overline{abc}\)

=1001\(\overline{abc}\)

=143.7.\(\overline{abc}\)

=> \(\overline{abcabc}\)

Ta có: 3a87+9a3+1a= 3000+a00+87+900+a0+3+10+a

=(3000+87+900+3+10) + ( a.100+ a.10+a)=4000+ a.(100+10+1)

=4000+a.111(1)

aaa+4000= a.111+4000(2)

Từ (1) và (2) suy ra: 3a87+9a3+1a= aaa+4000

Tick cho mik nha!!!

aaa + aa + a     = 861

111a + 11a + a = 861 (Cấu tạo số)

a(111+11+1)     = 861

123a                = 861

a                      = 861 : 123

a                      = 7

           Đảm bảo đúng 100%

aaa+aa+a=861

=> a00+a0+a+a0+a+a=861

=>a00+a0.2+a.3=861

=>a.100+a.10.2+a.3=861

=>a.100+a.20+a.3=861

=> a.(100+20+3)=861

=> a.123=861

=>a=861:123=7