1+ 9×10+81×72-71+811-8901+111111111111111111+111111111111
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{3}{2}+\dfrac{13}{6}+\dfrac{37}{12}+\dfrac{81}{20}+\dfrac{151}{30}+\dfrac{253}{42}+\dfrac{393}{56}+\dfrac{577}{72}+\dfrac{811}{90}\)
\(=\dfrac{459}{10}\)
Có:
+) \(81^4\equiv60\left(mod71\right)\)
\(\left(81^4\right)^2\equiv60^2\equiv50\left(mod71\right)\) (1)
+) \(27^5\equiv20\left(mod71\right)\)
\(\left(27^5\right)^2\equiv20^2\equiv45\left(mod71\right)\) (2)
+) \(9^7\equiv54\left(mod71\right)\)
\(\left(9^7\right)^2\equiv54^2\equiv5\left(mod71\right)\) (3)
Từ (1), (2), (3):
\(\Rightarrow81^8-27^{10}-9^{14}\equiv50-45-5\equiv0\left(mod71\right)\)
=> \(81^8-27^{10}-9^{14}⋮71\left(đpcm\right)\)
\(=\left(3^4\right)^8-\left(3^3\right)^{10}-\left(3^2\right)^{14}\)
\(=3^{32}-3^{30}-3^{28}\)
\(=3^{28}.\left(3^4-3^2-1\right)\)
\(=3^{28}.71_{ }\)
=> \(81^8-27^{10}-9^{14}\) chia hết cho 71
Đặt \(A=71^9+71^8+...+71^2+71+1\)
\(\Rightarrow71A=71^{10}+71^9+...+71^2+71\)
\(\Leftrightarrow70A=71^9-1\)
hay \(A=\dfrac{71^9-1}{70}\)
\(C=70\cdot A+1\)
\(=71^9-1+1=71^9\)
Đặt B=719+718+717+...+712+71
71B=7110+719+718+717+...+712
71B-B=7110-71
70B=7110-71=>B=\(\frac{71^{10}-71}{70}\)
Ta có A=70.\(\frac{71^{10}-71}{70}\)
=7110-71
1.1111122e+17