Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(\frac{1}{4}+-\frac{5}{13}\right)+\left(\frac{2}{11}+-\frac{8}{13}+\frac{3}{4}\right)\)
\(=\frac{2}{11}+\left(\frac{1}{4}+\frac{3}{4}\right)+\left(-\frac{5}{13}+-\frac{8}{13}\right)\)
\(=\frac{2}{11}+1+\left(-1\right)\)
\(=\frac{2}{11}+0\)
\(=\frac{2}{11}\)
b) \(\left(\frac{21}{31}+-\frac{16}{7}\right)+\left(\frac{44}{53}+\frac{10}{31}\right)+\frac{9}{53}\)
\(=-\frac{16}{7}+\left(\frac{21}{31}+\frac{10}{31}\right)+\left(\frac{44}{53}+\frac{9}{53}\right)\)
\(=-\frac{16}{7}+1+1\)
\(=-\frac{16}{7}+2\)
=\(-\frac{2}{7}\)
c) \(\frac{\frac{9}{45}-\frac{4}{13}-\frac{1}{3}}{\frac{3}{13}-\frac{1}{15}+\frac{2}{3}}\)
\(=-\frac{43}{81}\)
\(\frac{-7}{31}\) và \(\frac{6}{31}\)
\(\frac{-7}{31}<0;\frac{6}{31}>0\)
=>\(-\frac{7}{31}<\frac{6}{31}\)
\(\frac{-97}{128}\) và \(-\frac{99}{128}\)
vì \(\frac{97}{128}<\frac{99}{128}\) =>\(\frac{-97}{128}>-\frac{99}{128}\)
\(\frac37\) và \(\frac{-6}{7}\)
vì\(\frac37>0;-\frac67<0\)
=>\(\frac37>-\frac67\)
\(a,(\frac{1}{4}+\frac{-5}{13})+(\frac{2}{11}+\frac{-8}{13}+\frac{3}{4})\)
\(= (\frac{1}{4} + \frac{3}{4}) + (\frac{-5}{13} + \frac{-8}{13}) + \frac{2}{11}\)
\(= \frac{4}{4} + \frac{-13}{13} + \frac{2}{11}\)
\(=1+(-1)+\frac{2}{11}=\frac{2}{11}\)
\(b,(\frac{21}{31}+\frac{-16}{7})+(\frac{44}{53}+\frac{10}{31})+\frac{9}{53}\)
\(= (\frac{21}{31} + \frac{10}{31}) + (\frac{44}{53} + \frac{9}{53}) + \frac{-16}{7}\)
\(=\frac{31}{31}+\frac{53}{53}+\frac{-16}{7}=1+1-\frac{16}{7}\)
\(=2-\frac{16}{7}=\frac{14}{7}-\frac{16}{7}=-\frac27\)
\(c,\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)
\(= (\frac{-5}{7} + \frac{-2}{7}) + (\frac{3}{4} + \frac{1}{4}) + \frac{-1}{5}\)
\(= \frac{-7}{7} + \frac{4}{4} + \frac{-1}{5}\)
\(=-1+1-\frac{1}{5}=0-\frac15=-\frac15\)
\(\frac{-3}{31} + \frac{-6}{17} + \frac{1}{25} + \frac{-28}{31} + \frac{-11}{17} + \frac{-1}{5}\)
\(= (\frac{-3}{31} + \frac{-28}{31}) + (\frac{-6}{17} + \frac{-11}{17}) + \frac{1}{25} + \frac{-1}{5}\)
\(= \frac{-31}{31} + \frac{-17}{17} + \frac{1}{25} - \frac{5}{25}\)
\(= -1 + (-1) + \frac{-4}{25}\)
\(=-2-\frac{4}{25}=-\frac{50}{25}-\frac{4}{25}=-\frac{54}{25}\)
49\(\frac{8}{23}\) - (5\(\frac{7}{32}\) + 14\(\frac{8}{23}\))
= 49\(\frac{8}{23}\) - 5\(\frac{7}{32}\) - 14\(\frac{8}{23}\)
= (49\(\frac{8}{23}\) - 14\(\frac{8}{23}\)) - 5\(\frac{7}{32}\)
= 35 - 5 - \(\frac{7}{32}\)
= 30 - \(\frac{7}{32}\)
= \(\frac{960}{32}\) - \(\frac{7}{32}\)
= 953/32
Câu b:
71\(\frac{38}{45}\) - (43\(\frac{8}{45}\) - 1\(\frac{17}{57}\))
= 71\(\frac{38}{45}\) - 43\(\frac{8}{45}\) - 1\(\frac{17}{57}\)
= 28\(\frac{30}{45}\) - 1\(\frac{17}{57}\)
= 28\(\frac23\) - 1\(\frac{17}{57}\)
= \(\frac{86}{3}\) - \(\frac{74}{57}\)
= \(\frac{1634}{57}\) - \(\frac{74}{57}\)
= \(\frac{1560}{57}\)
= \(\frac{520}{19}\)
\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
\(C=\frac{16}{15.31}+\frac{14}{31.45}+\frac{7}{45.52}+\frac{7}{52.65}+\frac{1}{13.70}\)
\(C=\frac{16}{15.31}+\frac{14}{31.45}+\frac{7}{45.52}+\frac{13}{52.65}+\frac{5}{67.70}\)
\(C=\frac{1}{15}-\frac{1}{31}+\frac{1}{31}-\frac{1}{45}+\frac{1}{45}-\frac{1}{52}+\frac{1}{52}-\frac{1}{65}+\frac{1}{65}-\frac{1}{70}\)
\(C=\frac{1}{15}-\frac{1}{70}\)
\(C=\frac{11}{210}\)
Vậy: \(C=\frac{11}{210}\)
C \(=\frac{898}{17745}\)
Tk mk nhé
ban co the giai ra ko
cam on wrecking ball