\(\dfrac{4}{1x5}\) \(+\) \(\dfrac{4}{5x9}\) \(+\) \(\dfrac{4}{9x13}\) \(+\) ... \(+\) \(\dfrac{4}{53x57}\)
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7 tháng 10 2021
\(I=\dfrac{2}{1\times5}+\dfrac{2}{5\times9}+\dfrac{2}{9\times13}+...+\dfrac{2}{181\times185}\)
\(=\dfrac{1}{2}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{181\times185}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{181}-\dfrac{1}{185}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{185}\right)=\dfrac{1}{2}\times\dfrac{184}{185}=\dfrac{92}{185}\)
10 tháng 10 2021
\(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+...+\dfrac{1}{45\times49}\)
\(=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{45\times49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{49}\right)=\dfrac{1}{4}\times\dfrac{48}{49}=\dfrac{12}{49}\)
15 tháng 10 2021
\(\Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\left(x\ne0\right)\\ \Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\Rightarrow\dfrac{21}{3x}=\dfrac{21}{45}\Rightarrow3x=45\\ \Rightarrow x=15\)
NT
2
6 tháng 1 2022
\(S=\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{41\times45}\)
\(\Rightarrow4S=\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{41\times45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{8}{45}\)
\(\Rightarrow S=\dfrac{2}{45}\)
6 tháng 1 2022
\(=\dfrac{1}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{8}{45}=\dfrac{2}{45}\)
PL
0
31 tháng 10 2023
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)