1+1+1+1+1+1x5=
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Ta có: \(\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot10}+...+\dfrac{1}{2010\cdot2015}\)
\(=\dfrac{1}{5}+\dfrac{1}{5}\left(\dfrac{5}{5\cdot10}+\dfrac{5}{10\cdot15}+...+\dfrac{5}{2010\cdot2015}\right)\)
\(=\dfrac{1}{5}+\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{2010}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{5}+\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{402}{2015}\)
\(=\dfrac{1}{5}\left(1+\dfrac{402}{2015}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{2417}{2015}=\dfrac{2417}{10075}\)
\(B=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{125\times129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{125}-\dfrac{1}{129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{129}\right)=\dfrac{1}{4}\times\dfrac{128}{129}=\dfrac{32}{129}\)
\(D=\dfrac{1}{1\cdot5}+\dfrac{1}{5\cdot9}+...+\dfrac{1}{21\cdot25}\)
\(4D=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{21\cdot25}\)
\(4D=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{21}-\dfrac{1}{25}\)
\(4D=1-\dfrac{1}{25}=\dfrac{24}{25}\)
\(D=\dfrac{24}{25}\cdot\dfrac{1}{4}=\dfrac{4\cdot6}{25\cdot4}=\dfrac{6}{25}\)
\(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+...+\dfrac{1}{45\times49}\)
\(=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{45\times49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{49}\right)=\dfrac{1}{4}\times\dfrac{48}{49}=\dfrac{12}{49}\)
10 , ủng hô mk nha
1+1+1+1+1+1×5= 1+1+1+1+1+5 =10 nha bạn! Với lại lớp 1 chưa học nhân nên cho đây là toán lớp 2