\(\dfrac{24\cdot47-23}{24+47\cdot23}\cdot\dfrac{3+\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{7}-\dfrac{9}{11}+9}\\ =\dfrac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\dfrac{3\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}{9\left(1+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{1001}-\dfrac{1}{13}\right)}\\ =\dfrac{24\cdot24+24\cdot23-23\cdot1}{24+24\cdot23+23\cdot23}\cdot\dfrac{1}{3}\\ =\dfrac{23\left(24-1\right)+24\cdot24}{24\left(1+23\right)+23\cdot23}\cdot\dfrac{1}{3}=\dfrac{23\cdot23+24\cdot24}{24\cdot24+23\cdot23}\cdot\dfrac{1}{3}\\ =1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)

b, B = 1 + 2 + 2^2 + 2^3 +.....+ 2^2013

2B = 2.(1 + 2 + 2^2 + 2^3 +.....+ 2^2013)

2B = 2 + 2^2 + 2^3 + 2^4 +.....+ 2^2014

2B - B = 2^2014 - 1

B = 2^2014 - 1

a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)

\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)

\(=\dfrac{15}{26}-\dfrac{4}{26}\)

\(=\dfrac{11}{26}\)

b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)

\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)

\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)

\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)

\(=\dfrac{-967}{63}\)

c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)

\(=-1\)

d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)

=0

a: =27/45-20/45=7/45

b: \(=\dfrac{3}{5}+\dfrac{30}{40}=\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{12}{20}+\dfrac{15}{20}=\dfrac{27}{20}\)

c: \(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)

d: \(=\dfrac{9}{23}\left(\dfrac{5}{17}-\dfrac{22}{17}\right)+11+\dfrac{9}{23}=11\)

a) \(\dfrac{3}{5}+\dfrac{-4}{9}=\dfrac{27}{45}+\dfrac{-20}{45}=\dfrac{7}{45}\)

b) \(\dfrac{3}{5}+\dfrac{2}{5}.\dfrac{15}{8}=1.\dfrac{15}{8}=\dfrac{15}{8}\)

c) \(\dfrac{7}{2}.\dfrac{8}{13}+\dfrac{8}{13}.\dfrac{-5}{2}+\dfrac{8}{13}=\dfrac{8}{13}.\left(\dfrac{7}{2}+\dfrac{-5}{2}\right)=\dfrac{8}{13}.1=\dfrac{8}{13}\)

d) \(\dfrac{-5}{17}.\dfrac{-9}{23}+\dfrac{9}{23}.\dfrac{-22}{17}+11\dfrac{9}{23}=\dfrac{9}{23}.\left(\dfrac{-5}{17}+\dfrac{-22}{17}\right)=\dfrac{-243}{391}\)

a) \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\\ =\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\\ =\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{15}{26}-\dfrac{2}{13}\right)\\ =0+\left(\dfrac{15}{26}-\dfrac{4}{26}\right)\\ =0+\dfrac{11}{26}\\ =\dfrac{11}{26}\)

\(c)\dfrac{-11}{23}.\dfrac{6}{7}+\dfrac{8}{7}.\dfrac{-11}{23}-\dfrac{1}{23}\\=\dfrac{-1}{23}\left ( \dfrac{66}{7}+\dfrac{88}{7}+1 \right )\\ =\dfrac{-1}{23}.23=-1\)

\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=\dfrac{13}{15}\)

31​−53​+75​−97​+119​−1311​+1513​+1311​−119​+97​−75​+53​−31​
=(13−13)−(35−35)+(57−57)−(79−79)+(911−911)−(1113−1113)+1315=(31​−31​)−(53​−53​)+(75​−75​)−(97​−97​)+(119​−119​)−(1311​−1311​)+1513​
=1315=1513​

 

a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

A=0+0+0+...+0+\(\dfrac{13}{15}\)

A=\(\dfrac{13}{15}\)

b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)

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

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

\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)

\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)

\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)

\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)

\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)