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\(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+...+\dfrac{1}{45\times49}\)
\(=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{45\times49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{49}\right)=\dfrac{1}{4}\times\dfrac{48}{49}=\dfrac{12}{49}\)
\(S=\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{41\times45}\)
\(\Rightarrow4S=\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{41\times45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{8}{45}\)
\(\Rightarrow S=\dfrac{2}{45}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{8}{45}=\dfrac{2}{45}\)
\(\Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\left(x\ne0\right)\\ \Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\Rightarrow\dfrac{21}{3x}=\dfrac{21}{45}\Rightarrow3x=45\\ \Rightarrow x=15\)
`2/[1xx5]+2/[5xx9]+2/[9xx13]+....+2/[93xx97]+2/[97xx101]`
`=1/2xx(4/[1xx5]+4/[5xx9]+4/[9xx13]+....+4/[93xx97]+4/[97xx101])`
`=1/2xx(1-1/5+1/5-1/9+1/9-1/13+...+1/93-1/97+1/97-1/101)`
`=1/2xx(1-1/101)`
`=1/2xx100/101`
`=50/101`
\(B=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{125\times129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{125}-\dfrac{1}{129}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{129}\right)=\dfrac{1}{4}\times\dfrac{128}{129}=\dfrac{32}{129}\)
a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)
\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)
\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)\)
\(=2.\dfrac{99}{100}\)
\(=\dfrac{99}{50}\)
_____
b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)
\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)
\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=3.\left(1-\dfrac{1}{101}\right)\)
\(=3.\dfrac{100}{101}\)
\(=\dfrac{300}{101}\)
dấu " . " là dấu nhân nha bn, nãy viết nhanh quên lớp
3/(1×5) + 3/(5×9) + 3/(9×13) + 3/(13×17) + 3/(17×21)
= 3/4 × (1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + 1/17 - 1/21)
= 3/4 × (1 - 1/21)
= 3/4 × 20/21
= 5/7
\(\dfrac{4\times x}{1\times5}\) + \(\dfrac{4\times x}{5\times9}\) + \(\dfrac{4\times x}{9\times13}\) + \(\dfrac{4\times x}{13\times17}\) = 16
\(x\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+\dfrac{4}{13\times17}\right)\) = 16
\(x\) \(\times\) (\(\dfrac{1}{1}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{13}\) + \(\dfrac{1}{13}\) - \(\dfrac{1}{17}\)) = 16
\(x\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{17}\)) = 16
\(x\) \(\times\) \(\dfrac{16}{17}\) = 16
\(x\) = 16 : \(\dfrac{16}{17}\)
\(x\) = 17
\(I=\dfrac{2}{1\times5}+\dfrac{2}{5\times9}+\dfrac{2}{9\times13}+...+\dfrac{2}{181\times185}\)
\(=\dfrac{1}{2}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{181\times185}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{181}-\dfrac{1}{185}\right)\)
\(=\dfrac{1}{2}\times\left(1-\dfrac{1}{185}\right)=\dfrac{1}{2}\times\dfrac{184}{185}=\dfrac{92}{185}\)