\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)\)

\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}-\frac{23}{44}\)

\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)

\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)

\(=\frac{35}{35}+\left(\frac{-44}{44}\right)+\frac{2}{97}=1+\left(-1\right)+\frac{2}{97}=\frac{2}{97}\)

\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\) 

\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\) 

\(=-1+1+\frac{2}{97}\) 

\(=\frac{2}{97}\) 

=[3/5+(-3/7)-1/35]+[3/11-3/4-(-23/44)]+2/97

=[21/35+(-15/35)-1/35]+[12/44-33/44-(-23/44)+2/97

=(-1)+1+2/97

=0+2/97

=2/97

\(=\dfrac{3}{5}+\dfrac{3}{11}+\dfrac{3}{7}-\dfrac{1}{35}-\dfrac{3}{4}+\dfrac{-23}{44}\)

\(=\left(\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{1}{35}\right)+\left(\dfrac{3}{11}-\dfrac{3}{4}-\dfrac{23}{44}\right)\)

\(=\dfrac{21+15-1}{35}+\dfrac{12-33-23}{44}\)

\(=1-1=0\)

m) (\(\frac{-5}{12}\)+\(\frac{6}{11}\))+(\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\))

= \(\frac{-5}{12}\)+\(\frac{6}{11}\)+\(\frac{7}{17}\)+\(\frac{5}{11}\)+\(\frac{5}{12}\)

= (\(\frac{-5}{12}\)+\(\frac{5}{12}\))+(\(\frac{6}{11}\)+\(\frac{5}{11}\))+\(\frac{7}{17}\)

= 0+1+\(\frac{7}{17}\)

= \(\frac{24}{17}\)

n) (\(\frac{9}{16}\)+\(\frac{8}{-27}\))+(1+\(\frac{7}{16}\)+\(\frac{-19}{27}\))

= \(\frac{9}{16}\)+\(\frac{8}{-27}\)+1+\(\frac{7}{16}\)+\(\frac{-19}{27}\)

= (\(\frac{9}{16}\)+\(\frac{7}{16}\))+(\(\frac{8}{-27}\)+\(\frac{-19}{27}\))+1

= 1+(-1)+1

= 0+1

= 1

o) (6-2\(\frac{4}{5}\)).3\(\frac{1}{8}\)-1\(\frac{3}{5}\):\(\frac{1}{4}\)

= (6-\(\frac{14}{5}\)).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)

= \(\frac{16}{5}\).\(\frac{25}{8}\)-\(\frac{8}{5}\):\(\frac{1}{4}\)

= 10-\(\frac{8}{5}\):\(\frac{1}{4}\)

= 10-\(\frac{32}{5}\)

= \(\frac{18}{5}\)

CHÚC BẠN HỌC TỐT

\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\) _  \(\frac{3-\frac{3}{11}-\frac{3}{17}}{5-\frac{5}{11}-\frac{5}{17}}\)

=\(\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}\)_ \(\frac{3\left(1-\frac{1}{11}-\frac{1}{17}\right)}{5\left(1-\frac{1}{11}-\frac{1}{17}\right)}\)= \(\frac{2}{4}-\frac{3}{5}\)= \(\frac{-1}{10}\)

Ta có \(\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}=\left(\frac{3}{10}-\frac{6}{20}-\frac{4}{15}-\frac{1}{20}\right).\frac{5}{19}=-\left(\frac{4}{15}+\frac{1}{20}\right).\frac{5}{19}\)

\(=-\frac{1}{5}\left(\frac{4}{3}+\frac{1}{4}\right).\frac{5}{19}=-\left(\frac{4}{3}+\frac{1}{4}\right).\frac{1}{19}=-\frac{19}{12}.\frac{1}{19}=-\frac{1}{12}\)

Lại có \(\left(\frac{1}{14}+\frac{1}{7}-\frac{3}{35}\right).\frac{-4}{3}=\left(\frac{1}{14}+\frac{1}{7}-\frac{5}{35}+\frac{2}{35}\right).\frac{-4}{3}=\left(\frac{1}{14}+\frac{2}{35}\right).\frac{-4}{3}\)

\(=\frac{1}{7}\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{3}=\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{21}=\frac{7}{10}.\frac{-4}{21}=\frac{-2}{15}\)

Khi đó \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{3}{35}\right).\frac{-4}{3}}=\frac{\frac{1}{12}}{-\frac{2}{15}}=-\frac{5}{8}\)

Làm lại : 

\(\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\left(-\frac{4}{3}\right)=\left(\frac{1}{14}+\frac{1}{7}-\frac{5}{35}+\frac{2}{35}\right)\left(-\frac{4}{3}\right)=\left(\frac{1}{14}+\frac{2}{35}\right)\left(-\frac{4}{3}\right)\)

\(=\frac{1}{7}\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{3}=\frac{-4}{21}.\frac{9}{10}=\frac{-6}{35}\)

Khi đó A = \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{-\frac{1}{12}}{-\frac{6}{35}}=\frac{35}{72}\)

ta có :

\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{\frac{1}{5}\left(\frac{3}{2}-\frac{2}{3}-\frac{7}{4}\right).\frac{5}{19}}{\frac{1}{7}\left(\frac{1}{2}+1+\frac{3}{5}\right).\frac{-4}{3}}=\frac{\frac{1}{19}\left(-\frac{2}{3}-\frac{1}{4}\right)}{-\frac{3}{21}\left(\frac{3}{2}+\frac{3}{5}\right)}=\frac{\frac{1}{19}.\frac{-11}{12}}{-\frac{3}{21}.\frac{21}{10}}\)

\(=\frac{\frac{1}{19}.\frac{-11}{12}}{-\frac{3}{10}}=\frac{11}{19.12}.\frac{10}{3}=\frac{55}{342}\)