1: \(\frac{-5}{7}\cdot\frac{2}{11}+\frac{-5}{7}\cdot\frac{9}{11}+1\frac57\)

\(=-\frac57\left(\frac{2}{11}+\frac{9}{11}\right)+1+\frac57\)

\(=-\frac57+1+\frac57=1\)

2: \(-3\frac{4}{13}\cdot15\frac{3}{41}+3\frac{4}{13}\cdot2\frac{3}{41}\)

\(=3\frac{4}{13}\left(-15-\frac{3}{41}+2+\frac{3}{41}\right)\)

\(=\frac{43}{13}\cdot\left(-13\right)=-43\)

3: \(\frac45\cdot15\frac14-\frac45\cdot15\frac13+\frac{11}{30}\)

\(=\frac45\left(15+\frac14-15-\frac13\right)+\frac{11}{30}\)

\(=\frac45\cdot\left(\frac14-\frac13\right)+\frac{11}{30}\)

\(=\frac45\cdot\frac{-1}{12}+\frac{11}{30}=\frac{-1}{15}+\frac{11}{30}=\frac{-2}{30}+\frac{11}{30}=\frac{9}{30}=\frac{3}{10}\)

4: \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}-\frac35+\frac{2}{21}-\frac{10}{21}+\frac{3}{10}\)

\(=\left(\frac15+\frac25-\frac35\right)+\left(\frac{8}{21}+\frac{2}{21}-\frac{10}{21}\right)+\frac{3}{10}\)

=0+0+\(\frac{3}{10}=\frac{3}{10}\)

1

a)103/35

b)-3/13

a: \(\frac{-1}{9}\cdot\frac{-3}{5}+\frac{5}{-6}\cdot\frac{-3}{5}-\frac72\cdot\frac35\)

\(=\frac19\cdot\frac35+\frac56\cdot\frac35-\frac72\cdot\frac35\)

\(=\frac35\left(\frac19+\frac56-\frac72\right)\)

\(=\frac35\left(\frac{2}{18}+\frac{15}{18}-\frac{63}{18}\right)=\frac35\cdot\frac{-46}{18}=\frac35\cdot\frac{-23}{9}=\frac{-23}{3\cdot5}=-\frac{23}{15}\)

b: Sửa đề: \(-\frac37\cdot\frac{15}{13}-\frac37\cdot\frac{11}{13}-\frac37\)

\(=-\frac37\left(\frac{15}{13}+\frac{11}{13}\right)-\frac37\)

\(=-\frac37\cdot2-\frac37=-\frac37\cdot3=-\frac97\)

c,

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

= \(\dfrac{5}{9}.1\)

= \(\dfrac{5}{9}\)

a,

= \(\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{-4}{3}\)

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

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

= \(\dfrac{-2}{3}\)

d) 

` (-3)/7 + 5/13 + (-4)/7 =   -3/7 + (-4/7 )  + 5/13 = -1 + 5/13 = -8/13`

e2) 

`-5/21 + (-2)/21 + (-8)/4 = -7/21 + -8/4 = -7/3`

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

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

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​

 

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