Ta có :

\(1abc.2=abc8\)

\(\Leftrightarrow\left(1000+abc\right).2=10.abc+8\)

\(\Leftrightarrow2000+2.abc=10.abc+8\)

\(\Leftrightarrow10.abc-2.abc=10.abc+8\)

\(\Leftrightarrow10.abc-2.abc=2000-8\)

\(\Leftrightarrow8.abc=1992\)

\(\Leftrightarrow abc=249\)

Vậy số \(abc\) cần tìm là \(249\)

Ấn gì vậy chả hiểu

Giải:

Ta có:

\(\overline{1abc}.2=\overline{abc8}\)

\(\Rightarrow\left(1000+\overline{abc}\right).2=10.\overline{abc}+8\)

\(\Rightarrow2000+2.\overline{abc}=10.\overline{abc}+8\)

\(\Rightarrow10.\overline{abc}-2.\overline{abc}=2000-8\)

\(\Rightarrow8.\overline{abc}=1992\)

\(\Rightarrow\overline{abc}=249\)

\(\Rightarrow a=2,b=4,c=9\)

Vậy a = 2, b = 4, c = 9

Ta có:

1abc x 2 = abc8

=> (1000 + abc) x 2 = abc0 + 8

=> 2000 + abc x 2 = abc x 10 + 8

=> 2000 - 8 = abc x 10 - abc x 2

=> 1992 = abc x 8

=> abc = 1992 : 8

=> abc = 249

Vậy a = 2; b = 4; c = 9

1abc. 2 =abc8

(1000+abc) .2 = (abc.10)+8

2000+abc.2    = abc .10+80

abc.10-abc.2  =  2000-80

abc.(10-2)      =  1920

abc . 8          =  1920 

abc               =   1920 : 8

abc               =  240

abc không tồn tại.

Ủng hộ nhé.

K mk nhé!

bn lên mạng là có ngay!

K mk nhé!

k mk !

thanks!

Thank you!

Nhớ k mk!

#embengaytho#

Bài 1:

a)

\(\overline{abcd}=100\overline{ab}+\overline{cd}\)

\(=100.2\overline{cd}+\overline{cd}\)

\(=201\overline{cd}\)

Mà \(201⋮67\)

\(\Rightarrow\overline{abcd}⋮67\)

b)

\(\overline{abc}=100\overline{a}+10\overline{b}+\overline{c}\)

\(=\left(100\overline{b}+10\overline{c}+\overline{a}\right)+\left(99\overline{a}-90\overline{b}-9\overline{c}\right)\)

\(=\overline{bca}+9\left[\left(12\overline{a}-9\overline{b}\right)-\left(\overline{a}+\overline{b}+\overline{c}\right)\right]\)

\(=\overline{bca}+27\left(4\overline{a}-3\overline{b}\right)-\left(\overline{a}+\overline{b}+\overline{c}\right)⋮27\)

\(\Rightarrow\overline{bca}-\left(\overline{a}+\overline{b}+\overline{c}\right)⋮27\)

\(\Rightarrow\left\{{}\begin{matrix}\overline{bca}⋮27\\\overline{a}+\overline{b}+\overline{c}⋮27\end{matrix}\right.\)

\(\Rightarrow\overline{bca}⋮27\)

Bài 2:

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

\(=\overline{ab}.99+\overline{ab}+\overline{cd}\)

\(=\overline{ab}.11.99+\left(\overline{ab}+\overline{cd}\right)\)

Mà \(11⋮11\)

\(\Rightarrow\overline{ab}.11.9⋮11\)

\(\Rightarrow\overline{abcd}⋮11\).

Các bạn giải nhanh cho mình nhé. Thanks!