1.

\(\overline{aabb}=11\overline{a0b}\)

\(\Rightarrow1100a+11b=11\left(100a+b\right)\)

\(\Rightarrow1100a+11b=1100a+11b\)

\(\RightarrowĐCPM\)

2.

\(\overline{aaabbb}=111\overline{a00b}\)

\(\Rightarrow111000a+11b=111\left(100a+b\right)\)

\(\Rightarrow111000a+11b=111000a+11b\)

\(\RightarrowĐPCM\)

\(\overline{aabb}=100\overline{aa}+\overline{bb}=100a11+11b\)

\(=11.\left(100a+b\right)=\overline{a0b}.11\)

\(\overline{aaabbb}=1000a111+111b=111\left(1000a+b\right)\)

\(=\overline{a00b}.111\)

a, abab = 101ab 

Ta có: ab.101 = ab . ( 100 + 1 ) = ab00 + ab = abab

b, aaabbb = a00b . 111

Ta có: a00b.111 = a00b . 100 + a00b . 10 + a00b  = a00b00 + a00b0 + a00b = aa0bb0 + a00b = aaabbb  

a) ab.101=ab.(100+1)=ab.100+ab=(10a+b).100+10a+b=1000a+100b+10a+b=abab

b) a00b.111=a00b.(100+1)=a00b.100+a00b.10+a00b=a00b00+a00b0+a00b=aa00bb0=a00b=aaabbb

mk cũng đang hỏi giống bạn đây!

a)Ta có: Xét vế trái :

ab.101 <=> ab.(100+1) => ab.100+ab.1

=> ab00+ab=abab=Vế phải (đpcm)

b)Ta có Xét vế phải:

a00b.111=a00b.(100+10+1)=a00b00+a00b0+a00b

= aaabbb=Vế Phải(đpcm)

xong rui

ta có aaabbb = a. 10000 + a.1000 + a.100 + b .100 + b.10 + b.1

= a.1000 . ( 100 + 10 +1) + b . ( 100+10+1 )

=a.1000.111 + b.111

= 111.(a.1000 + b )

= 111.a00b

=> aaabbb = a00b .111

mk làm đúng chưa mina

18.(x-16) = 18

      x - 16 = 1

      x        = 17

\(\frac{aaabbb}{a00b}\)

\(=\frac{111000a+111b}{1000a+b}\)

\(=\frac{111\left(1000a+b\right)}{1000a+b}\)

\(=111\)