Bài 2:

- \(\frac75\) + \(x\) = - \(\frac34\)

\(x\) = - \(\frac34+\frac75\)

\(x=-\frac{15}{20}\) + \(\frac{28}{20}\)

\(x\) = \(\frac{13}{20}\)

Vậy \(x=\frac{13}{20}\)

Bài 3:

- \(\frac74\) + \(\frac14:x\) = \(\frac32\)

\(\frac14:x\) = \(\frac32+\frac74\)

\(\frac14:x\) = \(\frac64+\frac74\)

\(\frac14:x=\frac{13}{4}\)

\(x=\frac14:\frac{13}{4}\)

\(x=\frac14\times\frac{4}{13}\)

\(x=\frac{1}{13}\)

Vậy \(x=\frac{1}{13}\)

a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)

\(=\frac{8+9+10}{12}\)

\(=\frac{27}{12}=\frac{9}{4}\)

b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)

\(=\frac{45-14+20}{24}\)

\(=\frac{51}{24}=\frac{17}{8}\)

2)

a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)

\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)

\(=1+\frac{7}{13}+\frac{1}{7}\)

\(=\frac{20}{13}+\frac{1}{7}\)

\(=\frac{153}{91}\)

Tí tớ trả lời tiếp

b)\(5\frac14+3\frac25-4\frac14\)

=\(\left(5\frac14-4\frac14\right)+3\frac25\)

=\(\left\lbrack\left(5-4\right)+\left(\frac14-\frac14\right)\right\rbrack+\frac{17}{5}\)

=\(1+0+\frac{17}{5}\)

=\(\frac55+\frac{17}{5}\)

=\(\frac{22}{5}\)

a)\(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{117}{12}-\frac{2}{12}=\frac{115}{12}\)

b)\(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)

c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)

a. \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)

b. \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-1\frac{1}{2}\)

= \(\frac{13}{4}.1-\frac{3}{2}=\frac{13}{4}-\frac{3}{2}=\frac{7}{4}\)

c. \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2004}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2003}{2004}=\frac{1}{2004}\)

kazuto kirigaya thật là bt làm ko đó ko bt thì nói đi còn bt thì làm đi

trời ơi bài dễ thế này tự làm đi còn hỏi

\(a,(\frac{1}{4}+\frac{-5}{13})+(\frac{2}{11}+\frac{-8}{13}+\frac{3}{4})\)

\(= (\frac{1}{4} + \frac{3}{4}) + (\frac{-5}{13} + \frac{-8}{13}) + \frac{2}{11}\)

\(= \frac{4}{4} + \frac{-13}{13} + \frac{2}{11}\)

\(=1+(-1)+\frac{2}{11}=\frac{2}{11}\)

\(b,(\frac{21}{31}+\frac{-16}{7})+(\frac{44}{53}+\frac{10}{31})+\frac{9}{53}\)

\(= (\frac{21}{31} + \frac{10}{31}) + (\frac{44}{53} + \frac{9}{53}) + \frac{-16}{7}\)

\(=\frac{31}{31}+\frac{53}{53}+\frac{-16}{7}=1+1-\frac{16}{7}\)

\(=2-\frac{16}{7}=\frac{14}{7}-\frac{16}{7}=-\frac27\)

\(c,\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)

\(= (\frac{-5}{7} + \frac{-2}{7}) + (\frac{3}{4} + \frac{1}{4}) + \frac{-1}{5}\)

\(= \frac{-7}{7} + \frac{4}{4} + \frac{-1}{5}\)

\(=-1+1-\frac{1}{5}=0-\frac15=-\frac15\)

\(\frac{-3}{31} + \frac{-6}{17} + \frac{1}{25} + \frac{-28}{31} + \frac{-11}{17} + \frac{-1}{5}\)

\(= (\frac{-3}{31} + \frac{-28}{31}) + (\frac{-6}{17} + \frac{-11}{17}) + \frac{1}{25} + \frac{-1}{5}\)

\(= \frac{-31}{31} + \frac{-17}{17} + \frac{1}{25} - \frac{5}{25}\)

\(= -1 + (-1) + \frac{-4}{25}\)

\(=-2-\frac{4}{25}=-\frac{50}{25}-\frac{4}{25}=-\frac{54}{25}\)