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a; \(\dfrac{3}{5}\) - \(\dfrac{-7}{10}\) - \(\dfrac{13}{-20}\)
= \(\dfrac{12}{20}\) + \(\dfrac{14}{20}\) + \(\dfrac{13}{20}\)
= \(\dfrac{39}{20}\)
b; \(\dfrac{1}{2}\) + \(\dfrac{1}{-3}\) + \(\dfrac{1}{4}\) - \(\dfrac{-1}{6}\)
= \(\dfrac{6}{12}\) - \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{2}{12}\)
= \(\dfrac{7}{12}\)
Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+\cdots+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)\cdot n}=\frac{1}{n-1}-\frac{1}{n}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n-1}-\frac{1}{n}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac{1}{n}<1\)
=>\(\frac14\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)<\frac14\)
=>\(A<\frac14\) (ĐPCM)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(=2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2019-2018}{2018.2019}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2019}\right)\)
\(=\frac{2017}{2019}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2003\cdot2004}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2003}-\frac{1}{2004}\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+....\left(\frac{1}{2003}-\frac{1}{2003}\right)-\frac{1}{2004}\)
\(=1-0+0+0+....+0-\frac{1}{2004}\)
\(=1-\frac{1}{2004}\)
\(=\frac{2003}{2004}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(=1-\frac{1}{2004}\)
\(=\frac{2003}{2004}\)
`S=1/2 +1/6 +1/12 +1/20 +...+1/380`
`=1/(1.2)+1/(2.3) +1/(3.4)+1/(4.5)+...+1/(19.20)`
`=1-1/2 +1/2 -1/3 +1/3-1/4 +1/4 -1/5+....+1/19-1/20`
`=1-1/20=20/20 -1/20 =19/20`