Điều cần chưng minh là sai

Ví dụ: \(\overline{abcd}=7920⋮99\) nhưng \(79-20⋮̸99\) 

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!

Ta có : \(\overline{abcd}=10\overline{ab}+\overline{cd}=100.2.\overline{cd}+\overline{cd}\)

                    \(=201.\overline{cd}\)

Mà      \(201⋮67\)nên \(201.\overline{cd}⋮67\)

Vậy \(\overline{abcd}⋮67\)

Ta có: abcd = ab x 100 + cd =200cd +cd (vì ab = 2cd)

hay=201cd

Mà \(201⋮67\left(=3\right)\)

\(\Rightarrow201\overline{cd}⋮67\)

Vậy \(\overline{ab}=2\overline{cd}\Leftrightarrow\overline{abcd}⋮67\)

Ta có: \(\overline{abcdeg}=10000\overline{ab}+100\overline{cd}+\overline{eg}=9999\overline{ab}+99\overline{cd}+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)⋮11\)

b, 10²⁸+8 chia hết cho 9

10²⁸+8=(10²⁷*10)+8=1000⁹+8 chia hết cho 8

(8,9)=1 nên 10²⁸+8 chia hết cho 27

Ta có \(\dfrac{ }{abcd}=10.\dfrac{ }{ab}+\dfrac{ }{cd}=4.25.\dfrac{ }{ab}+\dfrac{ }{cd}\)

a) Nếu \(\dfrac{ }{cd}⋮4\Rightarrow\dfrac{ }{abcd}⋮4\)

b) Nếu \(\dfrac{ }{abcd}⋮4\) thì \(4.25.\dfrac{ }{ab}+\dfrac{ }{cd}⋮4\) nên \(\dfrac{ }{cd}⋮4\)

a) abcd = abx100 + cd = abx4x25 + cd
Vì abx4x25 chia hết cho 4;cd chia hết cho 4(gt)
=> abx4x25 + cd chia hết cho 4
=> abcd chia hết cho 4 (đpcm)
b) abcd = abx100 + cd = abx4x25 + cd
Vì abx4x25 chia hết cho 4;abcd chia hết cho 4 (gt)
=> cd chia hết cho 4 (đpcm)