\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\)

=>\(2S=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\)

=>\(2S-S=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{2020}}\)

=>\(S=1-\dfrac{1}{2^{2020}}< 1\)

\(S=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{1001\cdot1003}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{1001}-\dfrac{1}{1003}\)

\(=1-\dfrac{1}{1003}< 1\)

a: Những tia trên hình vẽ là Ex,Ey,Em,En,Ct,CK,Cn

Đoạn thẳng: EK,EC,CK

b: Các cặp tia đối nhau là:

Ex;Ey

Kx;Ky

Cn;CE

CK,Ct

\(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2023}\)

Đặt: \(C=\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2023}\)

\(\dfrac{3}{2}C=\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+...+\left(\dfrac{3}{2}\right)^{2024}\)

\(\dfrac{3}{2}C-C=\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+...+\left(\dfrac{3}{2}\right)^{2024}-\dfrac{3}{2}-\left(\dfrac{3}{2}\right)^2-...-\left(\dfrac{3}{2}\right)^{2023}\)

\(\dfrac{1}{2}C=\left(\dfrac{3}{2}\right)^{2024}-\dfrac{3}{2}\) 

\(C=2\left(\dfrac{3}{2}\right)^{2024}-3\)

\(\Rightarrow A=\dfrac{1}{2}+2\left(\dfrac{3}{2}\right)^{2024}-3\)

\(=2\left(\dfrac{3}{2}\right)^{2024}-\dfrac{5}{2}\)

\(\Rightarrow A-B=2\left(\dfrac{3}{2}\right)^{2024}-\dfrac{5}{2}-2\left(\dfrac{3}{2}\right)^{2024}=-\dfrac{5}{2}\)

Xác suất thực nghiệm không phải mặt 4 chấm là:

\(\dfrac{40-13}{40}=\dfrac{27}{40}\)

a) O nằm giữa A và B nên:

\(AB=OA+OB\)

\(\Rightarrow OA=1,5+3=4,5\left(cm\right)\)

b) C nằm giữa O và B 

\(OB=OC+BC\)

\(\Rightarrow BC=OB-OC\)

\(\Rightarrow BC=3-1,5=1,5\left(cm\right)\)

\(OC=BC=1,5\left(cm\right)\)