\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{18\times19}+\frac{2}{19\times20}\)
Tính nhanh
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
19 tháng 8 2015
\(\frac{1}{2\times2}+\frac{1}{3\times3}+....+\frac{1}{20\times20}
28 tháng 8 2017
Cho biểu thức A= 11×2×3 + 12×3×4 + 13×
4×5 +...+ 118×19×20 . So sánh A với 14 .
Dương Đình Hưởng
cố lên mà k
HT
31 tháng 7 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{19.20}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}\)
= \(1-\frac{1}{20}\)
= \(\frac{19}{20}\)
18 tháng 4 2017
\(\frac{3}{1^2x2^2}\)+\(\frac{5}{2^2x3^2}\)+...+\(\frac{39}{19^2x20^2}\)<1
=\(\frac{3}{1.4}\)+\(\frac{5}{4x9}\)+...+\(\frac{39}{361x400}\)<1
=1-\(\frac{1}{4}\)+\(\frac{1}{4}\)-...-\(\frac{1}{361}\)+\(\frac{1}{361}\)-\(\frac{1}{400}\)<1
vì 1-\(\frac{1}{400}\)<1 nên \(\frac{3}{1^2x2^2}\)+\(\frac{5}{2^2x3^2}\)+...+\(\frac{39}{39^2x40^2}\)<1
vậy..............................................
19 tháng 4 2017
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
18 tháng 10 2016
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
DD
2
MT
25 tháng 6 2015
\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{9\times10}\)
=\(2\times\frac{1}{1\times2}+2\times\frac{1}{2\times3}+2\times\frac{1}{3\times4}+...+2\times\frac{1}{9\times10}\)
=\(2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\right)\)
=\(2\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(2\times\left(\frac{1}{1}-\frac{1}{10}\right)=2\times\left(\frac{10}{10}-\frac{1}{10}\right)=2\times\frac{9}{10}\)
=\(\frac{9}{5}\)
25 tháng 6 2015
=2-1+1-\(\frac{2}{3}\)+\(\frac{2}{3}\)-\(\frac{1}{2}\)+...+\(\frac{2}{9}\)-\(\frac{1}{5}\)
=2-\(\frac{1}{5}\)
=\(\frac{10}{5}\)-\(\frac{1}{5}\)
=\(\frac{9}{5}\).
**** mình nha mấy bạn.
TM
29 tháng 4 2017
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)\)
9 tháng 11 2014
\(\frac{1}{8}=12,5\%\) ; \(\frac{1}{16}=6,25\%\) ; \(\frac{1}{2}=50\%\) ; \(\frac{1}{4}=25\%\)
Thay vào trên mà tính.
= \(1+\left(\frac{3\left(1x2+2x4x2\right)}{3\left(5+5x3x25\right)}+1\right)-\left(1+\frac{18}{54}\right)-1\) = \(\frac{18}{380}-\frac{18}{54}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+.......+\frac{2}{18.19}+\frac{2}{19.20}.\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)=\frac{2.19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...-\frac{1}{19}+\frac{1}{20}\)
\(=1+\frac{1}{20}\)
\(=\frac{1}{20}\)