Chứng minh rằng:
a.\(\overline{abcabc}\)\(⋮7\)
b. \(\overline{aaa}\) \(⋮37\)
c. \(\overline{1ab1}\) \(-\overline{1ba1}\) \(⋮90\)
chứng tỏ rằng
\(a,\overline{aaaaa}:7\)
\(b,\overline{aaa}:7\)
\(c,\overline{abcabc}:11\)
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.\(\overline{abc}\)
=> \(\overline{abcabc}\)
chứng minh rằng a) \(\overline{abcabc}\) chia hết cho 7, 11, 13
b) \(\overline{ab}-\overline{ba}\) chia hết cho 9
c) \(\overline{abc}-\overline{cba}\) chia hết cho 99
a) Ta có: \(\overline{abcabc}=100000a+10000b+1000c+100a+10b+c\) \(=100100a+10010b+1001c\) \(=1001\left(100a+10b+c\right)=7\cdot11\cdot13\left(100a+10b+c\right)⋮7,11,13\)
b) Ta có: \(\overline{ab}-\overline{ba}=10a+b-10b-a=9a-9b\) \(=9\left(a-b\right)⋮9\)
c) Ta có: \(\overline{abc}-\overline{cba}=100a+10b+c-100c-10b-a=99a-99c=99\left(a-c\right)⋮99\)
Tìm thương :
a) \(\overline{aaa}:a\)
b) \(\overline{abab}:\overline{ab}\)
c) \(\overline{abcabc}:\overline{abc}\)
a ) Ta có :
\(\overline{aaa}:a\)
\(=a.1.111:a.1\)
\(=111\)
b ) Ta có :
\(\overline{abab}:\overline{ab}\)
\(=\overline{ab}.100+\overline{ab}.1:\overline{ab}\)
\(=\overline{ab}.101:\overline{ab}\)
\(=101\)
c ) Ta có :
\(\overline{abcabc}:\overline{abc}\)
\(=\overline{abc}.1000+\overline{abc}.1:\overline{abc}\)
\(=\overline{abc}.1001:\overline{abc}\)
\(=1001\)
BT3 :Chứng Minh
1) \(\overline{aaa}\) \(⋮\) 37
2) \(\overline{aaa}\) + \(\overline{bbb}\) \(⋮\) 37
1)aaa=111a=37.3.a\(⋮37\)(đpcm)
2)aaa+bbb=111a+111b=111(a+b)\(⋮\)11(đpcm)
Dễ mà
Bài 1:Chứng minh rằng
a) \(\overline{ab}\) = 2.\(\overline{cd}\) → \(\overline{abcd}\) ⋮ 67
b) Cho \(\overline{abc⋮27}\) chứng minh rằng \(\overline{bca}\) ⋮ 27
Bài 2: Chứng minh rằng: Nếu \(\overline{ab}\) + \(\overline{cd}\) ⋮11 thì \(\overline{abcd}\) ⋮11
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!
Cho \(\overline{abc}\) ⋮ 37 . Chứng minh rằng \(\overline{cab}\) ⋮ 37
(abc) chia hết cho 37
->100.a + 10.b + c chia hết cho 37
-> 1000.a + 100.b + 10.c chia hết cho 37
-> 1000.a - 999.a + 100.b + 10.c chia hết cho 37 (vì 999.a chia hết cho 37)
-> 100.b + 10.c + a = (bca) chia hết cho 37 (bca) chia hết cho 37
-> 100.b+10.c+a chia hết cho 37
-> 1000.b + 100.c + 10.a chia hết cho 37
-> 1000.b - 999.b + 100.c + 10.a chia hết cho 37 (vì 999.b chia hết cho 37)
-> 100.c + 10.a + b = (cab) chia hết cho 37
Cho \(\dfrac{a+\overline{bc}}{\overline{abc}}=\dfrac{b+\overline{ca}}{\overline{bca}}=\dfrac{c+\overline{ab}}{\overline{cab}}\). Chứng minh rằng \(\dfrac{\overline{ab}}{c}=\dfrac{\overline{ca}}{b}=\dfrac{\overline{bc}}{a}\)
cho dãy tỉ số :
\(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}\) chứng minh rằng : a = b = c
Ta có:
\(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}\)
Mà: \(\left\{\begin{matrix}\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{10a+b+10b+c}{a+b}=9a+10b+c\\\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{10b+c+10c+a}{b+c}=9b+10c+a\\\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{10c+a+10a+b}{c+a}=9c+10a+b\end{matrix}\right.\)
\(\Rightarrow9a+10b+c=9b+10c+a=9c+10a+b\)
\(\Rightarrow\left\{\begin{matrix}9a=9b=9c\\10b=10c=10a\\c=a=b\end{matrix}\right.\)\(\Rightarrow a=b=c\)
Vậy \(a=b=c\) (Đpcm)
CHO BIẾT \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}\)
CHỨNG MINH RẰNG \(a=b=c\)
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}-\overline{bc}-\overline{ca}+\overline{ca}+\overline{ab}}{a+b-b-c+c+a}=\frac{2\overline{ab}}{2a}=10+\frac{b}{a}\)
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}+\overline{bc}+\overline{ca}-\overline{ca}-\overline{ab}}{a+b+b+c-c-a}=\frac{2\overline{bc}}{2b}=10+\frac{c}{b}\)
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{-\overline{ab}-\overline{bc}+\overline{bc}+\overline{ca}+\overline{ca}+\overline{ab}}{-a-b+b+c+c+a}=\frac{2\overline{ca}}{2c}=10+\frac{a}{c}\)
=> \(\frac{b}{a}=\frac{c}{b}=\frac{a}{c}\Rightarrow\frac{b+c+a}{a+b+c}=1\Rightarrow a=b=c\)