Ta có:

\(M=4^{10}+4^{11}+...+4^{199}\)

\(\Rightarrow M=\left(4^{10}+4^{11}\right)+...+\left(4^{198}+4^{199}\right)\)

\(\Rightarrow M=4^{10}\left(1+4\right)+...+4^{198}\left(1+4\right)\)

\(\Rightarrow M=4^{10}.5+...+4^{198}.5\)

\(\Rightarrow M=\left(4^{10}+...+4^{198}\right).5⋮5\)

\(\Rightarrow M⋮5\)

\(\Rightarrow M\in B\left(5\right)\left(đpcm\right)\)

ai tra loi se co qua

Ta có: M = 4^10+4^11+ ...+4^199 (tổng m có 190 số hạng)

M=(4^10+4^11)+(4^12+4^13)+ ... +(4^198+4^199)

M=4^10(1+4)+4^12(1+4)+4^198(1+4)

M=4^10*5+4^12*5+ ... +4^198*5

M=5(4^10+4^12+...+4^198)

Vì 5(4^10+4^12+...+4^198) chia hết cho 5 nên M chia hết cho 5

Quà đâu bạn

\(a\left(a+2\right)-a\left(a-5\right)-7=a^2+2a-a^2+5a-7\)

\(=\left(a^2-a^2\right)+\left(2a+5a\right)-7\)

\(=7a-7\)

\(=7\left(a-1\right)⋮7\) (đpcm)

\(a\left(a+2\right)-a\left(a-5\right)-7=a^2+2a-a^2-5a-7\)

\(=\) \(\left(a^2-a^2\right)-\left(2a+5a\right)-7\)

\(=7a-7\)

\(=7\left(a-1\right)\)\(⋮\)\(7\)( đpcm) 

\(a\left(a+2\right)-a\left(a-5\right)-7=a^2+2a-a^2-5a-7\)

\(=\) \(\left(a^2-a^2\right)-\left(2a+5a\right)-7\)

\(=7a-7\)

\(=7\left(a-1\right)\)\(⋮\)\(7\)( đpcm) 

Ta có: abcabc=abc.1001=abc.13.77 chia hết cho 77

=>abcabc chia hết cho 77

=>abcabc là bội của 77