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15: \(=\dfrac{7-13}{14}\cdot\dfrac{7}{5}-\dfrac{-2+3}{21}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\left(-\dfrac{6}{14}-\dfrac{1}{21}\right)\)
\(=\dfrac{7}{5}\cdot\dfrac{-10}{21}=\dfrac{-2}{3}\)
16: \(=\dfrac{3}{5}:\dfrac{-2-5}{30}+\dfrac{3}{5}:\left(\dfrac{-1}{3}-\dfrac{16}{15}\right)\)
\(=\dfrac{3}{5}\cdot\dfrac{-30}{7}+\dfrac{3}{5}:\dfrac{-5-16}{15}\)
\(=\dfrac{3}{5}\cdot\dfrac{-30}{7}+\dfrac{3}{5}\cdot\dfrac{-5}{7}\)
\(=\dfrac{3}{5}\cdot\left(-5\right)=-3\)
Đặt \(S=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1008}\right)\)
\(=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\)
Nên:
\(A=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)\(=\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right):\left(\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\)\(\Rightarrow A=1\)
Vậy A = 1
Chúc bạn học tốt!!
a) \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(=-\dfrac{1}{6}\cdot\dfrac{17}{28}\)
\(=-\dfrac{17}{168}\)
b) \(\left(\dfrac{15}{21}\div\dfrac{5}{7}\right)\div\left(\dfrac{6}{5}\div2\right)\)
\(=1\div\dfrac{3}{5}\)
\(=\dfrac{5}{3}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)