Tham khảo:

mình nghĩ kí hiệu đó là nhân 

Nếu là nhân thì lấy máy tính ra mà tính

Chúc bn hok tốt !

Kí hiệu * là gì vậy bn 

\(b=3.10^{100}+10^{99}+8=3.10^{100}+999...9+9⋮3\)

\(b=3.10^{100}+10^{99}+8⋮8\)

b đồng thời chia hết cho 3 và 8

3 và 8 nguyên tố cùng nhau và 3x8=24

=> b chia hết cho 24

\(A=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+....+\dfrac{\left(3^{99}+1\right)}{3^{99}}\)

\(A=\dfrac{4}{3}+\dfrac{10}{3^2}+\dfrac{28}{3^3}+...+\dfrac{\left(3^{99}+1\right)}{3^{99}}\)

\(A=\left(1+\dfrac{1}{3}\right)+\left(1+\dfrac{1}{3^2}\right)+\left(1+\dfrac{1}{3^3}\right)+...+\left(1+\dfrac{1}{3^{99}}\right)\)

\(A=\left(1+1+....+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)

\(A=99+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

Gọi \(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)là T

\(T=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

\(3T=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)

\(3T-T=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

\(2T=1-\dfrac{1}{3^{99}}\)

\(T=\left(1-\dfrac{1}{3^{99}}\right):2\)

\(T=\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}\)

\(=>A=99+T=99+\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}=99,5-\dfrac{1}{3^{99}\cdot2}< 100\)

Vậy A < 100

cảm ơn bn

a, \(4^{100}-4^{99}=4^{99}\left(4-1\right)=4^{99}\cdot3⋮3\)

vậy......

b, \(10^{15}+10^{16}+10^{17}=10^{15}\left(1+10+10^2\right)=10^5\cdot111⋮111\)

vậy.......