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
\(\Rightarrow A⋮30\)\(\Rightarrow A\in B\left(30\right)\)
a)Ta có:320=(32)10=910
230=(23)10=810
Vì 810<910
Suy ra:230<320
a)
\(P=\dfrac{5}{6}+\dfrac{5}{12}+\dfrac{5}{20}+\dfrac{5}{30}+\dfrac{5}{42}+\dfrac{5}{56}+\dfrac{5}{72}+\dfrac{5}{90}\\ =\dfrac{5}{2.3}+\dfrac{5}{3.4}+\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{5}{6.7}+\dfrac{5}{7.8}+\dfrac{5}{8.9}+\dfrac{5}{9.10}\\ \Rightarrow\dfrac{1}{5}P=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ =\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{5}{10}-\dfrac{1}{10}\\ =\dfrac{4}{10}=\dfrac{2}{5}\\ \Rightarrow P=\dfrac{2}{5}\cdot5=2\)
Vì tổng S có 100 SH
Mà 100 chia hết cho 2
Do đó ta có:
5+5^2+5^3+....+5^99+5^100
=(5+5^2)+(5^3+5^4)+...+(5^99+5^100)
=5.(1+5)+5^3.(1+5)+...+5^99.(1+5)
=5.6+5^3.6+...+5^99.6
=6.(5+5^3+...+5^99)
Vì 6 chia hết cho 6
Nên 6.(5+5^3+...+5^99) cũng chia hết cho 6
Vậy S chia hết cho 6
\(S=5+5^2+5^3+5^4+....+5^{99}+5^{100}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{99}+5^{100}\right)\)
\(=\left[5\left(1+5\right)\right]+\left[5^3\left(1+5\right)\right]+....+\left[5^{99}\left(1+5\right)\right]\)
\(=5\cdot6+5^3\cdot6+....+5^{99}\cdot6\)
\(=6\left(5+5^3+....+5^{99}\right)\)
\(\Rightarrow S⋮6\)
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 !!!!!!!!!!!!!!!!! 
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(4C=5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\)
\(4C-C=\left(5+\frac{5}{4}+...+\frac{5}{4^{98}}\right)-\left(\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\right)\)
\(3C=5-\frac{5}{4^{99}}\)
\(C=\frac{5-\frac{5}{4^{99}}}{3}\)
\(C=\frac{5}{3}-\frac{5}{4^{99}.3}< C\)
đpcm
55 - 54 + 53
= 53 ( 25 - 5 + 1 )
= 53. 21
Mà 21 ⋮ 7 ⇒ 55 - 54 + 53 ⋮ 7
\(A=5+5^2+5^3+5^4+...+5^{30}\)
\(\Rightarrow A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)
\(\Rightarrow A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{28}\left(5+5^2\right)\)
\(\Rightarrow A=\left(5+5^2\right)\left(1++5^2+...+5^{28}\right)\)
\(\Rightarrow A=30\left(1++5^2+...+5^{28}\right)⋮30\)