\(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\)

\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}\)

\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{9\times10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{10}{10}-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

=167,132

a: \(\dfrac{4}{6}+\dfrac{7}{13}+17+9+19+18+\dfrac{1}{9}+\dfrac{14}{6}\)

\(=\left(\dfrac{4}{6}+\dfrac{14}{6}\right)+\dfrac{1}{9}+\dfrac{7}{13}+63\)

\(=3+63+\dfrac{13+63}{117}\)

\(=66+\dfrac{76}{117}=\dfrac{7798}{117}\)

b: \(\dfrac{2}{5}\times\dfrac{1}{4}+\dfrac{3}{4}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{4}+\dfrac{3}{4}\right)=\dfrac{2}{5}\times1=\dfrac{2}{5}\)

b) \(\dfrac{2}{5}\times\dfrac{1}{4}+\dfrac{3}{4}\times\dfrac{2}{5}=\dfrac{2}{5}\times\left(\dfrac{1}{4}+\dfrac{3}{4}\right)=\dfrac{2}{5}\times1=\dfrac{2}{5}\)

2 lần vận tốc thật của cano là:

\(6\times5+8\times5=70\left(\dfrac{km}{h}\right)\)

vận tốc thật của cano là:

70:2=35(km/h)

Độ dài quãng sông AB là:

8x(35-5)=8x30=240(km)

\(\dfrac{3}{13}-\left(\dfrac{4}{7}+\dfrac{3}{13}\right)=\dfrac{3}{13}-\dfrac{4}{7}-\dfrac{3}{13}=\left(\dfrac{3}{13}-\dfrac{3}{13}\right)-\dfrac{4}{7}=0-\dfrac{4}{7}=-\dfrac{4}{7}\)

= 3/13 - 4/7 - 3/13
= (3/13 - 3/13) - 4/7
=0- 4/7
= -4/7

1: \(\dfrac{-5}{12}\cdot\dfrac{2}{11}+\dfrac{-5}{12}\cdot\dfrac{9}{11}+\dfrac{5}{12}\)

\(=-\dfrac{5}{12}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{5}{12}\)

\(=-\dfrac{5}{12}+\dfrac{5}{12}=0\)

2: \(\dfrac{-3}{5}:\dfrac{7}{5}+\dfrac{3}{5}:\dfrac{7}{5}+2\dfrac{3}{5}\)

\(=\left(-\dfrac{3}{5}+\dfrac{3}{5}\right):\dfrac{7}{5}+\dfrac{13}{5}\)

\(=0\cdot\dfrac{5}{7}+\dfrac{13}{5}=\dfrac{13}{5}\)

3: \(\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+\left(-2022\right)^0\)

\(=\dfrac{-3}{7}\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+1\)

\(=-\dfrac{3}{7}+1=\dfrac{4}{7}\)

4: \(0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2\cdot\dfrac{-1}{9}\)

\(=0,75-\dfrac{7}{3}-0,75+9\cdot\dfrac{-1}{9}\)

\(=-\dfrac{7}{3}-1=-\dfrac{10}{3}\)

5: \(2\dfrac{6}{7}\cdot\left[\left(-\dfrac{7}{5}-\dfrac{3}{2}:\dfrac{-5}{-4}\right)+\left(\dfrac{3}{2}\right)^2\right]\)

\(=\dfrac{20}{7}\cdot\left[-\dfrac{7}{5}-\dfrac{3}{2}\cdot\dfrac{4}{5}+\dfrac{9}{4}\right]\)

\(=\dfrac{20}{7}\left(-\dfrac{7}{5}-\dfrac{3}{10}+\dfrac{9}{4}\right)\)

\(=\dfrac{20}{7}\cdot\dfrac{-28-6+45}{20}\)

\(=\dfrac{45-34}{7}=\dfrac{11}{7}\)

6: \(\dfrac{2}{7}+\dfrac{5}{7}\left(\dfrac{3}{5}-0,25\right)\cdot\left(-2\right)^2+35\%\)

\(=\dfrac{2}{7}+\dfrac{5}{7}\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot4+\dfrac{7}{20}\)

\(=\dfrac{2}{7}+\dfrac{7}{20}+\dfrac{20}{7}\cdot\dfrac{7}{20}\)

\(=\dfrac{89}{140}+1=\dfrac{239}{140}\)

7: \(1\dfrac{13}{15}\cdot0,75-\left(\dfrac{11}{20}+25\%\right):1\dfrac{2}{5}\)

\(=\dfrac{28}{15}\cdot\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{5}{20}\right):\dfrac{7}{5}\)

\(=\dfrac{21}{15}-\dfrac{16}{20}\cdot\dfrac{5}{7}\)

\(=\dfrac{7}{5}-\dfrac{4}{7}=\dfrac{49-20}{35}=\dfrac{29}{35}\)

8: \(\left(-2,4+\dfrac{1}{3}\right):3\dfrac{1}{10}+75\%:1\dfrac{1}{2}\)

\(=\left(-\dfrac{12}{5}+\dfrac{1}{3}\right):\dfrac{31}{10}+\dfrac{3}{4}:\dfrac{3}{2}\)

\(=\dfrac{-31}{15}\cdot\dfrac{10}{31}+\dfrac{1}{2}\)

\(=-\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{-4+3}{6}=\dfrac{-1}{6}\)

a: 14,64-1,85-2,79

=14,64-(1,85+2,79)

=14,64-4,64

=10

b: 13,4m x5+13,4m x3+13,4m+13,4m

=13,4m x(5+3+1+1)

=13,4m x10

=134m

\(\left(-1,5\right)\cdot\dfrac{-2}{3}+\left(2,5-\dfrac{3}{4}\right):1\dfrac{3}{4}\)

\(=\dfrac{-3}{2}\cdot\dfrac{-2}{3}+\left(2,5-0,75\right):1,75\)

=1+1

=2