\(\overline{a0}\)\(.111=a0\) câu này theo mik phải là 11

nếu là 11:

\(\overline{a0}\)\(.11\)\(=\overline{aa0}\)

b)\(\overline{ab}\)\(.101=\overline{abab}\)

\(\overline{abc}\)\(.1001=abc000abc\)

abc *1001

=abc*(1000+1)

=abc*1000+abc*1

=abc000+ab

=ab000ab

A) ab.101=ab

B) abc. 7 .11.13=abcabc

k mình nha

a, ab.101=abab

b, abc.7.11.13=abc.1001=abcabc

ab . 101 = abab

abc . 7 . 11 . 13  = abc . 1001 = abcabc

a. ab.101=abab

b. abc.7.11.13=abc.1001=abcabc

pé pun

a﴿ ab x 101 = ab x ﴾100 + 1﴿

= ab x 100 + ab x 1

= ab00 + ab

= abab

b﴿ abc x 7 x 11 x 13

= abc x 1001

= abc x ﴾1000 + 1﴿

= abc x 1000 + abc x 1

= abc000 + abc

= abcabc

a) \(\overline{ab}.101=\overline{ab}.\left(100+1\right)\)

\(=\overline{ab}.100+\overline{ab}\)

\(=\overline{ab00}+\overline{ab}\)

\(=\overline{abab}\)

b) \(\overline{abc}.7.11.13=\overline{ab}.1001\)

\(=\overline{ab}.\left(1000+1\right)\)

\(=\overline{ab}.1000+\overline{ab}\)

\(=\overline{ab000}+\overline{ab}\)

\(=\overline{ab0ab}\)

a) abab

b) abcabc

 a / ab.101=ab(100+1) 
= ab.100+ab.1 
=ab00+ab 
=abab 
b/ abc.7.11.13=abc.1001 
=abc(1000+1) 
=abc.1000+abc.1 
=abc000+abc=abcabc

\(a,\overline{ab}\cdot101=\overline{abab}\)

\(b,\overline{abc}\cdot7\cdot11\cdot13=\overline{abc\cdot1001}=\overline{abcabc}\)

thank nha

1 .

a0 . 111 = aaa0

ab . 101 = abab 

abc . 1001 = abcabc

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rồi tớ làm bài 2

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