Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=5+5^2+5^3+...+5^8\)
\(A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^6\left(5+5^2\right)\)
\(A=30+5^2.30+...+5^6.30\)
Vì 30\(⋮\)30
\(\Rightarrow A⋮30\)\(\Rightarrow A\in B\left(30\right)\)
55 - 54 + 53
= 53 ( 25 - 5 + 1 )
= 53. 21
Mà 21 ⋮ 7 ⇒ 55 - 54 + 53 ⋮ 7
52+ 53 + 54 + ... + 510
= ( 52 + 53 ) + ( 54 + 55 ) + ... + ( 59 + 510 )
= 52.( 1 + 5 ) + 54.(1 + 5 ) + ... + 59.( 1 + 5 )
= 52.6 + 54.6 + ... + 59.6chia hết cho 6
Mà số chia hết cho 6 thì chia hết cho 3
Vậy tổng trên chia hết cho cả 3 và 6
5^2+5^3+5^4+...+5^9+5^10
=(5^2+5^3)+(5^4+5^5)+...+(5^9+5^10)
=(5^2.1+5^2.5)+(5^4.1+5^5.5)+...+(5^9.1+5^9.5)
=5^2.(1+5)+5^4.(1+5)+...+5^9.(1+5)
=5^2.6+5^4.6+...+5^9.6
=6.(5^2+5^4+...+5^9)
=2.3.(5^2+5^4+...+5^9)
Vậy tổng trên chia hết cho 3 và 6
Ta có: \(2^{n+3}+5^n-2^{n+1}+5^{n+1}=\left(2^{n+3}+2^{n+1}\right)+\left(5^n+5^{n+1}\right)\)
\(=2^n\left(2^3-2\right)+5^n\left(1+5\right)=2^n.6+5^n.6=6.\left(2^n+5^n\right)⋮6\left(đpcm\right)\)
Kb với mình nhé ~_~
2^n+3+5^n-2^n+1+5^n+1
= (2^n+3-2^n+1) + (5^n+5^n+1)
= 2^n.(2^3-2)+5^n.(5+1)
= 2^n.6+5^n.6 = 6.(2^n+5^n) chia hết cho 6
k mk nha
1) 55 - 54 + 53 = 53 . 52 - 53 . 5 - 53
= 53 . ( 52 - 5 + 1 )
= 53 . ( 25 - 5 - 1 )
= 53 . 21
= 53 . 3 . 7 chia hết cho 7
Vậy chứng minh 55 - 54 + 53 chia hết cho7
2) 76 + 75 - 74 = 74 . 72 + 74 . 7 - 74
= 74 . ( 72 + 7 - 1 )
= 74 . ( 49 + 7 - 1 )
= 74 . 55
= 74 . 5 .11 chia hết cho 11
Vậy chứng minh 76 + 75 - 74 chia hết cho 11
Tích mình nha !!!!!!!!!!!!!!!!! 
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
Bài 1:
a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)
\(=\frac{9}{15}+\frac{4}{15}\)
\(=\frac{13}{15}\)
b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)
\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)
c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)
\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)
d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)
\(=\frac{-7}{8}:\frac{-7}{4}\)
\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)
e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)
f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)
\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)
\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)
\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)
\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)
h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)
\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)
\(=\frac{5}{9}\)
Ta có 1 + 5 + 52 + 53 + ... + 520 + 521 (22 hạng tử)
= (1 + 5) + (52 + 53) + ... + (520 + 521) (11 cặp số)
= (1 + 5) + 52(1 + 5) + ... + 520(1 + 5)
= 6 + 52.6 + ... + 520.6
= 6(1 + 52 + ... + 520) \(⋮\)6 (đpcm)
1 + 5 + 52 + 53 + ... + 520 + 521
= ( 1 + 5 ) + ( 52 + 53 ) + ... + ( 520 + 521 )
= 6 + 52( 1 + 5 ) + ... + 520( 1 + 5 )
= 1.6 + 52.6 + ... + 520.6
= 6( 1 + 52 + ... + 520 ) chia hết cho 6 ( đpcm )