Thực hiện phép tính 5/ 1.3+ 5/3.5+ 5/5.7+...+5/ 99.101

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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