1) Tính
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
b)\(\left(\frac{15}{1\cdot2\cdot3}+\frac{15}{2\cdot3\cdot4}+\frac{15}{3\cdot4\cdot5}+.....+\frac{15}{18\cdot19\cdot20}\right)\cdot x=1\)
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BH
6 tháng 4 2018
Ta có: \(\frac{-3}{1.2.3}+\frac{-3}{2.3.4}+\frac{-3}{3.4.5}+...+\frac{-3}{18.19.20}\)
\(=\frac{-3}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}\right)\)
\(=\frac{-3}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{-3}{2}\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{-3}{2}.\frac{189}{380}=\frac{-567}{760}\)
DP
24 tháng 6 2017
2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\frac{2575}{5151}\)
\(=2,499514657\)
7 tháng 2 2017
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
23 tháng 12 2016
A=1/1x2 +1/2x3 +... +1/18x19 + 1/19x20
Nhận xét 1/1x2 = 1/1 -1/2 ; 1/2x3=1/2-1/3; ... ;1/18x19=1/18-1/19 ; 1/19x20=1/19-1/20
ta có A=1/1 - 1/2 + +1/2 -1/3+1/3- +1/18-1/19+1/19-1/20
A=1/1 - 1/20
A=20/20 - 1/20
A=(20-1)/20
A=19/20
Vậy A=19/20
23 tháng 12 2016
A =\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+ ...+\(\frac{1}{18.19}\)+\(\frac{1}{19.20}\)
A = 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+....+\(\frac{1}{18}\)- \(\frac{1}{19}\)+ \(\frac{1}{19}\)- \(\frac{1}{20}\)
A = 1 - \(\frac{1}{20}\)( Vì đã triệt tiêu )
A = \(\frac{19}{20}\)
9 tháng 10 2018
\(A=\frac{3^7\cdot17-3^9}{2^3\cdot3^5}=\frac{3^7\left(17-3^2\right)}{2^3\cdot3^5}=\frac{3^7\cdot2^3}{2^3\cdot3^5}=9\)
\(B=\frac{3^2\cdot4^2\cdot2^{32}}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{3^2\cdot2^{36}}{2^{35}\cdot11-2^{36}}=\frac{3^2\cdot2^{36}}{2^{35}\left(11-2\right)}=\frac{3^2\cdot2^{36}}{2^{35}\cdot3^2}=2\)
9 tháng 10 2018
\(\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2\cdot3^{28}}=\frac{3^{29}\cdot8}{2^2\cdot3^{28}}=6\)
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\cdot\frac{8}{33}\)
\(=\frac{52}{33}\)
a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99
A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)
A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)
A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
A= 13/2 ( 1/3 - 1/11)
A= 13/2 . 8/33
A= 52/33