\(\frac{1}{6}\)+
K
Khách

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3 tháng 3 2016

a)-2/3

b)4/39

c)26/45

3 tháng 3 2016

a) \(\frac{1}{6}+\frac{-5}{6}\)

\(=\frac{1+\left(-5\right)}{6}\)

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

b) \(\frac{6}{13}+\frac{-14}{39}\)

\(=\frac{18}{39}+\frac{-14}{39}\)

\(=\frac{4}{39}\)

c) \(\frac{4}{5}+\frac{4}{-18}\)

\(=\frac{4}{5}+\frac{-4}{18}\)

\(=\frac{72}{90}+\frac{-20}{90}\)

\(=\frac{52}{90}=\frac{26}{45}\)

10 tháng 1 2018

d) \(\frac{7}{14}+\frac{9}{36}\)

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

4) \(\frac{6}{7}=\frac{6.10}{7.10}=\frac{60}{70}\)

\(\frac{11}{10}=\frac{11.7}{10.7}=\frac{77}{70}\)

ta thay \(60< 77\)nen \(\frac{6}{7}< \frac{11}{10}\)

nhung cau khac lam tuong tu nhe 

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

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

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

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

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

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

27 tháng 2 2018

\(\frac{-5}{12}+\frac{-1}{4}=\frac{-5}{12}+\frac{-3}{12}=\frac{-5+-3}{12}=\frac{-8}{12}=\frac{-2}{3}\)

\(\frac{5}{12}+\frac{-3}{28}=\frac{35}{84}+\frac{-9}{84}=\frac{35+\left(-9\right)}{84}=\frac{26}{84}=\frac{13}{42}\)

\(\frac{-7}{15}+\frac{3}{35}=\frac{-49}{105}+\frac{9}{105}=\frac{-49+9}{105}=\frac{-40}{105}=\frac{-8}{21}\)

\(\frac{-5}{7}+\frac{-3}{4}=\frac{-20}{28}+\frac{-21}{28}=\frac{-20+\left(-21\right)}{28}=\frac{-41}{28}\)

16 tháng 4 2017

Giải bài 42 trang 26 SGK Toán 6 Tập 2 | Giải toán lớp 6

17 tháng 4 2017

a. \(\dfrac{-3}{5}\)

b. \(\dfrac{-2}{3}\) c. \(\dfrac{4}{39}\) d. \(\dfrac{26}{45}\)

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\frac{5}{19}+\frac{8}{43}\)

\(=\left(3\frac{14}{19}+6\frac{5}{19}\right)+\left(\frac{35}{43}+\frac{8}{43}\right)+\frac{13}{17}\)

\(=10+1+\frac{13}{17}=11+\frac{13}{17}=11\frac{13}{17}\)

\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)

\(=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1+\frac{-5}{7}.\left(-1\right)\)

\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}-1\right)+1\)

\(=\frac{-5}{7}.0+1==0+1=1\)