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16 tháng 9 2018
S = 1.3 + 3.5 + 5.7 + ...+ 99.101
=>6S = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 99.101.6
6S = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) + ...+ 99.101.(103-97)
6S = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 99.101.103 - 97.99.101
6S = 1.3 + 99.101.103
S = 171 650
16 tháng 9 2018
S = 1.3 + 3.5 + 5.7 + ...+ 99.101
=>6S = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 99.101.6
6S = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) + ...+ 99.101.(103-97)
6S = 1.3.5 + 1.3 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 99.101.103 - 97.99.101
6S = 1.3 + 99.101.103
S = 171 650
T
21 tháng 1 2017
đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}\)
ta có:
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}\)
=> 2A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{91.93}\)
=> 2A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{93}\)
=> 2A = \(\frac{1}{1}-\frac{1}{93}\)
2A = \(\frac{92}{93}\)
=> A = \(\frac{92}{93}:2\)
A = \(\frac{46}{93}\)
23 tháng 9 2018
Có: A=\(\dfrac{3}{1.5}+\dfrac{3}{5.10}+...+\dfrac{3}{100.105}\)
=> A=\(3.\dfrac{5}{5}\left(\dfrac{1}{1.5}+\dfrac{1}{5.10}+...+\dfrac{1}{100.105}\right)\)
=> A= \(3.\dfrac{1}{5}\left(\dfrac{5}{1.5}+\dfrac{5}{5.10}+...+\dfrac{5}{100.105}\right)\)
=> A=\(\dfrac{3}{5}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{105}\right)\)
=> A= \(\dfrac{3}{5}\left(1-\dfrac{1}{105}\right)\)=\(\dfrac{3}{5}.\dfrac{104}{105}=\dfrac{312}{525}\)
FM
23 tháng 9 2018
Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
NC
16 tháng 8 2020
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
Gọi A = 1.3+3.5+5.7+...+21.23
=> A = 1.(1+2)+3.(3+2)+5.(5+2)+...+21.(21+2)
=> A = 12+1.2+32+2.3+52+2.5+...+212+2.21
=> A = 12+32+52+...+212+(1.2+3.2+5.2+...+2.21)
Gọi B = 12+32+52+...+212
=> B = (21.22.23)/3
Gọi C = 1.2+2.3+5.2+...+2.21
=> C = 2(1+3+5+...+21)
=> C = 2{(21+1).[(21-1):2+1]}/2
=> C = 22x11=242
Vậy A = (21.22.23)/3+242