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A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
1)\(\dfrac{-5}{2}:\dfrac{1}{4}\) = \(\dfrac{-5}{2}\) x \(\dfrac{4}{1}\) = \(\dfrac{-20}{2}\)
2A=1-1/2+1/2^2-...+1/2^98-1/2^99
=>3A=1-1/2^100
=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)
\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)
\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)
\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)
\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)
\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)
\(E=\dfrac{3}{2}+\dfrac{1}{4}\)
\(E=\dfrac{6}{4}+\dfrac{1}{4}\)
\(E=\dfrac{7}{4}\)
\(P=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow\dfrac{1}{2}P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}\)
\(\Rightarrow\dfrac{1}{2}P-P=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{101}}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{100}}\)
\(\Rightarrow-\dfrac{1}{2}P=\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\)
\(\Rightarrow P=\left(\dfrac{1}{2^{101}}-\dfrac{1}{2^2}\right):\left(-\dfrac{1}{2}\right)\)
Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)
\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}\)
\(=\dfrac{2}{45}\)
-Chúc bạn học tốt-
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}\)
\(\Rightarrow2A-A=2-\dfrac{1}{2^{2012}}\)
\(\Rightarrow A=2-\dfrac{1}{2^{2012}}\)
\(A= 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\)\(\dfrac{1}{2^{2012}}\)
⇒\(2A=2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\)
⇒\(2A-A=(2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\))\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\right)\)
⇒\(A=2-\)\(\dfrac{1}{2^{2012}}\)