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\(\frac{2,3}{7}+\left(\frac{2}{9}-\frac{10}{7}\right)-\frac{5}{3}:\frac{1}{9}\)
\(\frac{2,3}{7}-\frac{76}{63}-15\)
\(-\frac{55,3}{63}-15\)
\(-15\frac{55,3}{63}\)
a. \(\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)+\frac{15}{26}-\frac{2}{13}\)
\(=\frac{15}{26}-\frac{2}{13}\)
\(=\frac{15}{26}-\frac{4}{26}=\frac{11}{26}\)
b. \(\frac{3}{7}-\frac{5}{3}\div\frac{1}{9}-\frac{3}{7}\)
\(=\frac{3}{7}-\frac{5}{3}\times\frac{9}{1}-\frac{3}{7}\)
\(=\frac{3}{7}-15-\frac{3}{7}\)
\(=-15\)
\(-\frac{3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\frac{15}{26}-\frac{2}{13}=\frac{11}{26}\)
\(\frac{2.3}{7}-\frac{5}{3}:\frac{1}{9}-\frac{3}{7}=\frac{6}{7}-15-\frac{3}{7}=\frac{3}{7}-15=-\frac{102}{7}\)
a, \(\dfrac{5}{1.3}\)+\(\dfrac{5}{3.5}\)+\(\dfrac{5}{5.7}\)+...+\(\dfrac{5}{99.101}\)
= 5.\(\dfrac{1}{1.3}\)+5.\(\dfrac{1}{3.5}\)+5.\(\dfrac{1}{5.7}\)+...+5.\(\dfrac{1}{99.101}\)
=5.(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{99.101}\))
=5.(\(\dfrac{2}{2}\).\(\dfrac{1}{1.3}\)+\(\dfrac{2}{2}\).\(\dfrac{1}{3.5}\)+\(\dfrac{2}{2}\).\(\dfrac{1}{5.7}\)+...+\(\dfrac{2}{2}\).\(\dfrac{1}{99.101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{99.101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{101}\))
=\(\dfrac{5}{2}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{101}\))
=\(\dfrac{5}{2}\).\(\dfrac{100}{101}\)
=\(\dfrac{250}{101}\)
=\(2\dfrac{48}{101}\)
b,\(\dfrac{-11}{23}\).\(\dfrac{6}{7}\)+\(\dfrac{8}{7}\).\(\dfrac{-11}{23}\)-\(\dfrac{1}{23}\)
=\(\dfrac{-11}{23}\).(\(\dfrac{6}{7}\)+\(\dfrac{8}{7}\))-\(\dfrac{1}{23}\)
=\(\dfrac{-11}{23}\).2-\(\dfrac{1}{23}\)
=\(\dfrac{-22}{23}\)-\(\dfrac{1}{23}\)
=-1
c,\(\dfrac{2.3}{7}\)+(\(\dfrac{2}{9}\)-\(1\dfrac{1}{3}\))-\(\dfrac{5}{3}\):\(\dfrac{1}{9}\)
=\(\dfrac{6}{7}\)+(\(\dfrac{2}{9}\)-\(\dfrac{4}{3}\))-\(\dfrac{5}{3}\).9
=\(\dfrac{6}{7}\)-\(\dfrac{10}{9}\)-\(\dfrac{5}{3}\).9
=\(\dfrac{6}{7}\)-\(\dfrac{10}{9}\)-15
=\(\dfrac{54}{63}\)-\(\dfrac{70}{63}\)-\(\dfrac{945}{63}\)
=\(\dfrac{-961}{63}\)=\(-15\dfrac{16}{63}\)
d,(20+\(9\dfrac{1}{4}\)):\(2\dfrac{1}{4}\)
=(20+\(\dfrac{37}{4}\)):\(\dfrac{9}{4}\)
=20:\(\dfrac{9}{4}\)+\(\dfrac{37}{4}\):\(\dfrac{9}{4}\)
=20.\(\dfrac{4}{9}\)+\(\dfrac{37}{4}\).\(\dfrac{4}{9}\)
=\(\dfrac{80}{9}\)+\(\dfrac{37}{9}\)
=\(\dfrac{117}{9}\)
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