Đặt \(A=\dfrac{10^{2022}+1}{10^{2023}+1};B=\dfrac{10^{2021}+1}{10^{2022}+1}\)

\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)

\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)

Vì \(10^{2023}+1>10^{2022}+1\)

nên \(\dfrac{9}{10^{2023}+1}< \dfrac{9}{10^{2022}+1}\)

=>\(\dfrac{9}{10^{2023}+1}+1< \dfrac{9}{10^{2022}+1}+1\)

=>10A<10B

=>A<B

\(\dfrac{5}{x+1}=\dfrac{20}{-12}\\ \Rightarrow\dfrac{5}{x+1}=\dfrac{5}{-3}\\ \Rightarrow x+1=-3\\ \Rightarrow x=-4\)

\(\dfrac{5}{x+1}=\dfrac{20}{-12}\)

\(\Rightarrow\dfrac{5}{x+1}=\dfrac{5}{-3}\)

\(\Rightarrow x+1=-3\)

\(\Leftrightarrow x=-4\)

8,1 + (-3,7) + 1,9 + (-6,3)

=[(-3,7) + (-6,3)] + (8,1 + 1,9)

=10 - 10

=0

\(\text{8,1 + (-3,7) + 1,9 + (-6,3) }\)

\(\text{=[(-3,7) + (-6,3)] + (8,1 + 1,9) }\)

\(=\left(-10\right)+10\)

\(=0\)

(-100)+78=-(100-78)=-22

(-100)+78=-(100-78)=-22

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2005}}\)

\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2004}}\)

\(3A-A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2005}}\right)\)

\(2A=1-\dfrac{1}{3^{2005}}\)

\(A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{2005}}\)

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2005}}\)

=>\(3A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2004}}\)

=>\(3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2004}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{2005}}\)

=>\(2A=1-\dfrac{1}{3^{2005}}=\dfrac{3^{2005}-1}{3^{2005}}\)

=>\(A=\dfrac{3^{2005}-1}{2\cdot3^{2005}}\)

\(-\dfrac{9}{25}\cdot17\dfrac{2}{3}-\left(-\dfrac{3}{5}\right)^2\cdot\dfrac{22}{3}\)

\(=-\dfrac{9}{25}\cdot\dfrac{53}{3}-\dfrac{9}{25}\cdot\dfrac{22}{3}\)

\(=-\dfrac{9}{25}\left(\dfrac{53}{3}+\dfrac{22}{3}\right)=-\dfrac{9}{25}\cdot25=-9\)

\(\dfrac{3}{7}\cdot\left(-\dfrac{2}{5}\right)\cdot2\dfrac{1}{2}\cdot20\cdot\dfrac{19}{72}\)

\(=\dfrac{3}{7}\cdot\left(-\dfrac{2}{5}\right)\cdot\dfrac{5}{2}\cdot20\cdot\dfrac{19}{72}\)

\(=\left(\dfrac{3}{7}\cdot\dfrac{19}{72}\right)\cdot\left(-\dfrac{2}{5}\cdot\dfrac{5}{2}\right)\cdot20\)

\(=\dfrac{19}{168}\cdot-1\cdot20\)

\(=\dfrac{19}{168}\cdot-20\)

\(=\dfrac{19\cdot-5}{42}\)

\(=\dfrac{-95}{42}\)

=-6/5

Đặt: \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)

\(A=1-\dfrac{1}{2^{2005}}\) 

Bài 8:

Tổng số: 50 lần

a) Số lần xuất hiện mặt 4 chấm: 12 lần 

Xác suất thực nghiệm ra được mặt 4 chấm là: 

\(P\left(A\right)=\dfrac{12}{50}=\dfrac{6}{25}\)

b) Tổng số lần xuất hiện số chấm lẻ là: `8+3+10=21` lần

Xác suất thực nghiệm ra được mặt số chấm lẻ là:

\(P\left(B\right)=\dfrac{21}{50}\) 

c) Tổng số lần xuất hiện số chấm nhỏ hơn 3 là: `8+7=15` lần

Xác suất thực nghiệm ra được số chấm nhỏ hơn 3 là:

\(P\left(C\right)=\dfrac{15}{50}=\dfrac{3}{10}\)