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Giải:
Đặt \(A=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\)
\(A=\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
\(A=\dfrac{1}{3}.\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)
\(A=\dfrac{1}{3}.\dfrac{9}{20}\)
\(A=\dfrac{3}{20}\)
\(\dfrac{1}{10}\)+\(\dfrac{1}{40}\)+\(\dfrac{1}{88}\)+\(\dfrac{1}{154}\)+\(\dfrac{1}{238}\)+\(\dfrac{1}{340}\)
=\(\dfrac{1}{2.5}\)+\(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+\(\dfrac{1}{14.17}\)+\(\dfrac{1}{17.20}\)
=\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{5}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{8}\)-\(\dfrac{1}{11}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{11}\)-\(\dfrac{1}{14}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{14}\)-\(\dfrac{1}{17}\))+\(\dfrac{1}{3}\).(\(\dfrac{1}{17}\)-\(\dfrac{1}{20}\))
=\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{14}\)+\(\dfrac{1}{14}\)-\(\dfrac{1}{17}\)+\(\dfrac{1}{17}\)+\(\dfrac{1}{20}\))
=\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\)+\(\dfrac{1}{20}\))
=\(\dfrac{1}{3}\).\(\dfrac{10+1}{20}\)
=\(\dfrac{1}{3}\).\(\dfrac{11}{20}\)
=\(\dfrac{11}{60}\)
Lời giải:
Gọi tử số là $T$
\(T=(1-\frac{1}{6})+(1-\frac{2}{7})+(1-\frac{3}{8})+....+(1-\frac{88}{93})\)
\(=\frac{5}{6}+\frac{5}{7}+\frac{5}{8}+....+\frac{5}{93}=5(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})\)
Gọi mẫu số là $M$
\(M=\frac{-1}{2}(\frac{1}{6}+\frac{1}{7}+....+\frac{1}{93})\)
Do đó:
\(C=\frac{5(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})}{\frac{-1}{2}(\frac{1}{6}+\frac{1}{7}+...+\frac{1}{93})}=\frac{5}{\frac{-1}{2}}=-10\)
Giải:
\(\dfrac{\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{2}{7}\right)+\left(1-\dfrac{3}{8}\right)+...+\left(1-\dfrac{88}{93}\right)}{\dfrac{-1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-...-\dfrac{1}{186}}\)
Gọi dãy là A,phần tử là B. Ta có:
B=\(\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{2}{7}\right)+\left(1-\dfrac{3}{8}\right)+...+\left(1-\dfrac{88}{93}\right)\)
B=\(\dfrac{5}{6}+\dfrac{5}{7}+\dfrac{5}{8}+...+\dfrac{5}{93}\)
B=5.\(\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{93}\right)\)
B=5.\(\left[\dfrac{2}{2}.\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{93}\right)\right]\)
B=5.\(\left[2.\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)\right]\)
B=10.\(\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)\)
⇒A=\(\dfrac{10.\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)}{\dfrac{-1}{12}+\dfrac{-1}{14}+\dfrac{-1}{16}+...+\dfrac{-1}{186}}\)
⇒A=-10
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\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7} +.....................+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+..........+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.......+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...........+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+...........+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.............+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+..........+\dfrac{1}{49.51}\right)}\)
\(=\dfrac{100}{2}\)
\(=50\)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+....+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+.....+\dfrac{1}{49.51}\right)}=\dfrac{\dfrac{100}{99}+\dfrac{100}{3.97}+....+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+....+\dfrac{1}{49.51}\right)}=\dfrac{100}{2}=50\)
e) \(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{68}}=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}{\dfrac{1}{4}\left(\dfrac{1}{2}-\dfrac{1}{13}+\dfrac{1}{17}\right)}=\dfrac{1}{3}:\dfrac{1}{4}=\dfrac{3}{4}\)
\(\dfrac{2}{40}+\dfrac{2}{88}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{202}{1540}\)
\(\Leftrightarrow2\left(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\right)=\dfrac{202}{1540}\)
\(\Leftrightarrow\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
a: \(=\dfrac{6}{18}+\dfrac{3}{18}+\dfrac{2}{18}+\dfrac{1}{18}=\dfrac{15}{18}=\dfrac{5}{6}\)
b: \(=\dfrac{7}{4}-\dfrac{5}{4}-\dfrac{2}{4}=0\)
c: \(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=1/2-1/10=4/10=2/5
Đặt A= \(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)
A= \(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\) 3A = \(3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+\dfrac{1}{14.17}+\dfrac{1}{17.20}\right)\) 3A= \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}\) 3A= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}\) 3A= \(\dfrac{1}{2}-\dfrac{1}{20}\)
3A= \(\dfrac{9}{20}\)
A= \(\dfrac{9}{20}:3\)
A = \(\dfrac{3}{20}\)
Vậy A= \(\dfrac{3}{20}\)
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