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\(\frac{1}{3}+\frac{13}{15}+\frac{33}{35}+...+\frac{9997}{9999}=1-\frac{2}{3}+1-\frac{2}{15}+1-\frac{2}{35}+...+1-\frac{2}{9999}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)
\(=50-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{4950}{101}\)
Mình giúp bạn nha!
A = 2017/1 + 2017/2 + 2017/3 + . . . + 2017/2018 / 2017/1 + 2016/2 + 2015/3 + . . .+ 1/2017
= 2017 . ( 1 + 1/2 + 1/3 + . . . +1/2018 ) / ( 2017 . 2016 . 2015 . . . 1) . ( 1 + 1/2 + 1/3 +. . . + 1/2017 )
= 1/2016 . 2015 . 2014. . . 1
k mình nha
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
=3018/5604
bạn ơi có thể nói rõ cách làm ko