a) = 3/3 x ( -24/54 +45/54 ) : 7/12

   = 1 x 21/54 x 12/7

   = 18/27 

( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )

S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)

S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)

S = \(\frac12\) - \(\frac19\)

S = \(\frac{9}{18}-\frac{2}{18}\)

S = \(\frac{7}{18}\)

\(\left(6+\left(\frac{1}{2}\right)^3-\left|-\frac{1}{2}\right|\right):\frac{3}{12}\)

\(=\left(6+\frac{1}{8}+\frac{1}{2}\right):\frac{1}{4}\)

=\(\frac{53}{8}:\frac{1}{4}\)

\(=\frac{53}{2}\)

a)30/60,-40/60,45/60,48/60

45/60>30/60>-40/60>-48/60

=3/4>1/2>-2/3>-4/5

Cộng các tổng ở các mẫu số được:    \(S=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}.\) 

       \(\Leftrightarrow S=1+\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{6}+\frac{1}{10}+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{21}+\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{36}.\) 

        Thực hiện các phép nhân một số với một hiệu ,được:

            \(S=1+\frac{1}{2}-\frac{1}{6}+\frac{1}{6}+\frac{1}{10}+\frac{1}{6}-\frac{1}{15}+\frac{1}{21}+\frac{1}{12}-\frac{1}{21}+\frac{1}{36}.\) 

         Giản ước, làm gọn được :   \(S=(1+\frac{1}{2})+(\frac{1}{10}+\frac{1}{6}-\frac{1}{15})+(\frac{1}{12}+\frac{1}{36}).\) 

            \(\Leftrightarrow S=\frac{3}{2}+\frac{1}{5}+\frac{1}{9}=\frac{135+18+10}{90}=\frac{163}{90}.\)

\(\frac{x}{5}=\frac23\)

\(x\) = \(\frac23\times5\)

\(x=\frac{10}{3}\)

Vậy \(x=\frac{10}{3}\)

\(\frac{x}{3}-\frac12=\frac15\)

\(\frac{x}{3}\) = \(\frac15\) + \(\frac12\)

\(\frac{x}{3}\) = \(\frac{2}{10}+\frac{5}{10}\)

\(\frac{x}{3}=\frac{7}{10}\)

\(x=\frac{7}{10}\times3\)

\(x=\frac{21}{10}\)

Vậy \(x=\frac{21}{10}\)

\(\frac{x}{5}+\frac12=\frac{6}{10}\)

\(\frac{x}{5}=\frac{6}{10}-\frac12\)

\(\frac{x}{5}=\frac{6}{10}-\frac{5}{10}\)

\(\frac{x}{5}=\frac{1}{10}\)

\(x=\frac{1}{10}\times5\)

\(x=\frac12\)

Vậy \(x=\frac12\)

\(\frac{x+3}{15}\) = \(\frac13\)

\(x+3=\frac13\times15\)

\(x+3=5\)

\(x=5-3\)

\(x=2\)

Vậy \(x=2\)

a)\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)

=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{14}{24}\right)\times\frac65+\frac12\)

=\(\frac{5}{24}\times\frac65+\frac12\)

=\(\frac14+\frac12\)

=\(\frac14+\frac24\)

=\(\frac34\)

b)\(\frac12+\frac34-\left(\frac34-\frac45\right)\)

=\(\frac12+\frac34-\frac34+\frac45\)

=\(\left(\frac12+\frac45\right)+\left(\frac34-\frac34\right)\)

=\(\left(\frac{5}{10}+\frac{8}{10}\right)+0\)

=\(\frac{13}{10}\)

a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)

... . . . .

\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)

\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)

b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

   \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

Suy ra \(\frac{2}{5}< S\) (1)

Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)

Từ đó suy ra S < 8/9

Từ (1) và (2) suy ra đpcm

a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)

\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)

\(=0-\frac{19}{26}\)

\(=-\frac{19}{26}\)

c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}.2-\frac{1}{23}\)

\(=\frac{-22}{23}-\frac{1}{23}\)

\(=-1\)

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

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

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

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

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