\(3,5+4\frac{1}{3}\)             b,\(8\frac{2}{3}-6,25...">
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24 tháng 7 2019

a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)

\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)

\(=1-\frac{4}{5}\)

\(=\frac{1}{5}\)

b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)

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

\(=2-1\)

\(=1\)

c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)

\(=\frac{-1}{4}+\frac{-16}{11}\)

\(=\frac{-75}{44}\)

d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)

\(=\frac{-6}{11}-\frac{3}{22}\)

\(=\frac{15}{22}\)

e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\) 

\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)

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

\(=\frac{1}{3}-\frac{1}{15}\)

\(=\frac{4}{15}\)

Bài 2: 

b: x+25%x=-1,25

=>1,25x=-1,25

hay x=-1

c: x-75%x=1/4

=>1/4x=1/4

hay x=1

Bài 2: 

a: =3/2-11/4=6/4-11/4=-5/4

b: =-49/6-17/2=-49/6-51/6=-100/6=-50/3

1 tháng 6 2020

sao bn phũ với mk thế:(( đx ko giải lại còn nói thếbucminh

3 tháng 6 2020

Nó dễ mà :(

1 tháng 8 2020

thanks friend!vui

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)

5 tháng 6 2020

Bài làm

a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)

\(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)

\(-\frac{3}{7}+\frac{7}{2}\)

\(=-\frac{7}{14}+\frac{49}{14}\)

\(=\frac{42}{14}=3\)

b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)

\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)

\(=5-\frac{1}{3}+\frac{2}{9}\)

\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)

\(=\frac{44}{9}\)

c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)

\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)

\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)

\(=\frac{5}{12}+\frac{7}{12}\)

\(=\frac{12}{12}=1\)

d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)

\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)

\(=-\frac{5}{9}+\frac{1}{90}\)

\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)

14 tháng 6 2016

a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)

                                                              \(=1-\frac{1}{32}=\frac{31}{32}\)

b) \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)\

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)