\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}=\dfrac{1}{9}-\dfrac{1}{10}\)

Do đó: \(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{9^2}< \dfrac{1}{8\cdot9}=\dfrac{1}{8}-\dfrac{1}{9}\)

Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}=\dfrac{8}{9}\)

Suy ra: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)

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

=>\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}\)

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

=>\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{99}}< \dfrac{1}{2}\)

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

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

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

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

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

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

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

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 

\(OB=OC+BC\)

\(\Rightarrow BC=OB-OC\)

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

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