Tìm số tự nhiên x thoả mãn 2x + 7 chia hết cho x + 1

Cho A=\(5\)+\(5^2\)+\(5^3\)+...+\(5^{2022}\) .Tìm x để 4A+5=\(5^x\)

Tính hợp lý:

A= 21.\(10^2\)-11.100+90.\(10^2\)+\(2^2\).\(5^2\).125.16

B= \(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)

C= 3+5-7-9+11+13 -13-17+19+...+2013-2015-2017+2019+2020

=2

=28

\(\dfrac{5}{13}\).\(\dfrac{-3}{10}\).\(\dfrac{-13}{5}\)

=\(\left(\dfrac{5}{13}.\dfrac{-13}{5}\right)\).\(\dfrac{-3}{10}\)

=\(\dfrac{-65}{65}\).\(\dfrac{-3}{10}\)

= -1.\(\dfrac{-3}{10}\)

=\(\dfrac{3}{10}\)