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A = \(\dfrac{3}{2}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{12}\) - \(\dfrac{9}{20}\) + \(\dfrac{11}{30}\) - \(\dfrac{13}{42}\) + \(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)
A = (1 + \(\dfrac{1}{2}\)) - (\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\)) + (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)) - (\(\dfrac{1}{4}\) + \(\dfrac{1}{5}\)) + (\(\dfrac{1}{5}\) + \(\dfrac{1}{6}\)) - (\(\dfrac{1}{6}\) + \(\dfrac{1}{7}\)) + (\(\dfrac{1}{7}\) + \(\dfrac{1}{8}\)) - (\(\dfrac{1}{8}\) + \(\dfrac{1}{9}\))
A = 1 + \(\dfrac{1}{2}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\)
A = 1 - \(\dfrac{1}{9}\)
A = \(\dfrac{8}{9}\)
\(A=\left(1+\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{7}+\dfrac{1}{8}\right)-\left(\dfrac{1}{8}+\dfrac{1}{9}\right)\)
\(A=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{8}-\dfrac{1}{9}\)
\(A=1+\dfrac{1}{9}=\dfrac{10}{9}\)
\(=\dfrac{15}{7\cdot8}-\dfrac{13}{6\cdot7}+\dfrac{11}{5\cdot6}-\dfrac{9}{4\cdot5}+\dfrac{7}{3\cdot4}-\dfrac{5}{2\cdot3}+\dfrac{3}{1\cdot2}\)
\(=\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{2}\)
=1+1/8=9/8
\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(B=2.\frac{100}{101}=\frac{200}{101}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
a)
\(P=\dfrac{5}{6}+\dfrac{5}{12}+\dfrac{5}{20}+\dfrac{5}{30}+\dfrac{5}{42}+\dfrac{5}{56}+\dfrac{5}{72}+\dfrac{5}{90}\\ =\dfrac{5}{2.3}+\dfrac{5}{3.4}+\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{5}{6.7}+\dfrac{5}{7.8}+\dfrac{5}{8.9}+\dfrac{5}{9.10}\\ \Rightarrow\dfrac{1}{5}P=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ =\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{5}{10}-\dfrac{1}{10}\\ =\dfrac{4}{10}=\dfrac{2}{5}\\ \Rightarrow P=\dfrac{2}{5}\cdot5=2\)
Trả lời
a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...\dfrac{2}{99.101}\)
=\(2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)
=\(2.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= \(2.\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
=\(2.\dfrac{100}{101}\)
=\(\dfrac{200}{101}\)
Ta có :
M= \(\dfrac{3+3-3+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4+4-4+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)= \(\dfrac{3+3-3}{4+4-4}=\dfrac{3}{4}\)
b) Nhận xét thấy: \(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};...\)
Ta có:
B= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
B= 1- \(\dfrac{1}{101}\)= \(\dfrac{100}{101}\)
Vậy B= \(\dfrac{100}{101}\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\\ =\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\\ =\dfrac{1}{2}+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\dfrac{4}{5}\\ =\dfrac{1}{2}+0+\dfrac{4}{5}\\ =\dfrac{1}{2}+\dfrac{4}{5}\\ =\dfrac{5}{10}+\dfrac{8}{10}\\ =\dfrac{13}{10}\)
\(\dfrac{-3}{7}+\dfrac{3}{4}:\dfrac{3}{14}\\ =\dfrac{-3}{7}+\dfrac{3}{4}\cdot\dfrac{14}{3}\\ =\dfrac{-3}{7}+\dfrac{7}{2}\\ =\dfrac{-6}{14}+\dfrac{49}{14}\\ =\dfrac{43}{14}\)
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)