Vận tốc của xe máy đi quãng đường dài 96km trong 2 giờ là:

\(96:2=48km\text{/}h\)

Với vận tốc đó đi hết quãng đường 120km trong thời gian là:

\(120:48=2,5h\)

Vậy ...

Thời gian xe máy đi tới B là:

\(8h-6h30'=1h30'=\dfrac{3}{2}h\)

Thời gian xe đạp tới B là:

\(1h30'+4h30'=6h\)

Vận tốc xe máy là:

\(67,5:\dfrac{3}{2}=45km\text{/}h\)

Vận tốc xe đạp là:

\(67,5:6=11,25km\text{/}h\)

Vậy ...

\(A=\dfrac{1}{2.3}+\dfrac{2}{3.5}+\dfrac{3}{5.8}+\dfrac{4}{8.12}+\dfrac{5}{12.17}+\dfrac{6}{17.23}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{23}\)

\(=\dfrac{1}{2}-\dfrac{1}{23}=\dfrac{21}{46}\)

Đặt \(\dfrac{1}{x+1}=a;\dfrac{1}{y-5}=b\)

\(\left\{{}\begin{matrix}a-b=1\\5a+3b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-3b=3\\5a+3b=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}\\b=-\dfrac{3}{4}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=\dfrac{11}{3}\end{matrix}\right.\)

\(C=\dfrac{14+\dfrac{7}{15}+\dfrac{7}{4}}{2+\dfrac{1}{15}+\dfrac{1}{4}}:\dfrac{5+\dfrac{5}{17}+\dfrac{5}{13}}{6+\dfrac{6}{17}+\dfrac{6}{13}}+\dfrac{5858}{5050}\)

\(=\dfrac{7\left(2+\dfrac{1}{15}+\dfrac{1}{4}\right)}{2+\dfrac{1}{15}+\dfrac{1}{4}}:\dfrac{5\left(1+\dfrac{1}{17}+\dfrac{1}{13}\right)}{6\left(1+\dfrac{1}{17}+\dfrac{1}{13}\right)}+\dfrac{58}{50}\)

\(=7:\dfrac{5}{6}+\dfrac{29}{25}=7.\dfrac{6}{5}+\dfrac{29}{25}\)

\(=\dfrac{42}{5}+\dfrac{29}{25}=\dfrac{239}{25}\)

\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right).....\left(1+\dfrac{1}{2020.2022}\right)\)

\(=\dfrac{1.3+1}{1.3}.\dfrac{2.4+1}{2.4}.\dfrac{3.5+1}{3.5}....\dfrac{2020.2022+1}{2020.2022}\)

\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{2021^2}{2020.2022}\)

\(=\dfrac{2.2021}{2022}=\dfrac{4042}{2022}\)

\(A=2+2^2+2^3+...+2^{2022}\)

\(2A=2^2+2^3+2^4+...+2^{2023}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2023}\right)-\left(2+2^2+2^3+...+2^{2022}\right)\)

\(A=2^{2023}-2\)