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24 tháng 6 2018

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-1-\frac{1}{2}-...-\frac{1}{5}\)

\(=\frac{1}{6}+\frac{1}{7}+...+\frac{1}{10}\left(đpcm\right)\)

24 tháng 6 2018

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\)\(\frac{1}{10}\)

\(A=\frac{1}{1}+\frac{1}{3}+...+\frac{1}{9}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{10}\)

\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}-2.\frac{1}{2}-2.\frac{1}{4}-...-2.\frac{1}{10}\)

\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}-1-\frac{1}{2}-...-\frac{1}{5}\)

\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\left(đpcm\right)\)

                                    ~~~Hok tốt~~~

                            

18 tháng 6 2017

Ta có : $16A=\dfrac{16}{6.10}+\dfrac{16}{7.9}+\dfrac{16}{8.8}+\dfrac{16}{9.7}+\dfrac{16}{10.6}$

$=>16A=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{6}$

$=>16A=2.(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$

$=>A=\dfrac{1}{8}(dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$

18 tháng 6 2017

\(A=\dfrac{1}{6.10}+\dfrac{1}{7.9}+\dfrac{1}{8.8}+\dfrac{1}{9.7}+\dfrac{1}{10.6}\)

\(16A=\dfrac{16}{6.10}+\dfrac{16}{7.9}+\dfrac{16}{8.8}+\dfrac{16}{9.7}+\dfrac{16}{10.6}\)
\(16A=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{6}\)

\(16A=2\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)

\(A=2:16\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)

\(A=\dfrac{1}{8}\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\left(đpcm\right)\)

19 tháng 5 2023

a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)

= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)

\(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)

= 1 - 1 - \(\dfrac{11}{25}\)

= - \(\dfrac{11}{25}\)

19 tháng 5 2023

b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)

= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))

= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)

= - \(\dfrac{499}{35}\)

10 tháng 4 2016

ta có:\(A=\frac{8^9+12}{8^9+7}=\frac{8^9+7+5}{8^9+7}=\frac{8^9+7}{8^9+7}+\frac{5}{8^9+7}=1+\frac{5}{8^9+7}\)

\(B=\frac{8^{10}+4}{8^{10}-1}=\frac{8^{10}-1+5}{8^{10}-1}=\frac{8^{10}-1}{8^{10}-1}+\frac{5}{8^{10}-1}=1+\frac{5}{8^{10}-1}\)

vì 810-1>89+7

\(\Rightarrow\frac{5}{8^{10}-1}<\frac{5}{8^9+7}\)

\(\Rightarrow1+\frac{5}{8^{10}-1}<1+\frac{5}{8^9+7}\)

=>A<B

12 tháng 5 2016

Chưa nghĩ ra...!!!