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NT
tính giá trị biểu thức
C=1/15+1/35+.................+1/2499
B=5/1.4+5/4.7+...................5/100.103
0
PV
11 tháng 4 2016
a) A x 3/5 = 3/1.4 + 3/4.7 + 3/7.10 + ...+ 3/100.103
= 1 - 1/4 + 1/4 - 1/7 + 1/7 -1/10 + ... + 1/100 - 1/103
= 1 - 1/103
= 102/103 => A = 102/103 : 3/5 = 170/103
b) B = 1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/49.51
B x2 = 2/3.5 + 2/5.7 + 2/7.9 + ...+ 2/49.51
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1.49 - 1/51
= 1/3 - 1/51
= 16/51
=> B = 16/51 : 2 = 8/51
6 tháng 5 2018
\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)
\(A=\frac{1}{2}.\frac{16}{51}\)
\(A=\frac{8}{51}\)
6 tháng 5 2018
\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\)
\(2A=\frac{1}{3}-\frac{1}{50}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{50}\right)\)
\(A=\frac{1}{2}.\frac{1}{3}-\frac{1}{2}.\frac{1}{50}\)
\(A=\frac{1}{6}-\frac{1}{100}=\frac{50}{300}-\frac{3}{300}=\frac{47}{300}\)
VT
15 tháng 4 2017
=1/2(2/3.5 + 2/5.7 +.....+2/49.51
=1/2(1/3 - 1/5+1/5-1/7+....+1/49-1/51)
=1/2(1/3-1/51)
=1/2.16/51
=8/51
HỌC TỐT NHÉ BẠN!
NL
1
LP
10 tháng 5 2015
a) 1/n-1/n+a=a/n.(n+a)
=(n+a)-n/n.(n+a)=n-n+a/n.(n+a)
=a/n.(n+a)
b)1/15+1/35+...+1/2499
=1/3.5+1/5.7+......+1/49.51
=1.2/2.3.5+1.2/2.5.7+.....+1.2/2.49.51
=1/2(2/3.5+2/5.7+.....+2/49.51)
=1/2(1/3-1/5+1/5-1/7+.......+1/49-1/50)
=1/2(1/3-1/50)
=1/2(50/150-3/150)
=1/2.47/150
=47/300
29 tháng 5 2017
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}.\dfrac{16}{51}=\dfrac{8}{51}\)
Vậy \(C=\dfrac{8}{51}\)
QD
29 tháng 5 2017
\(\Rightarrow C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(C=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(C=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(C=\dfrac{1}{2}.\dfrac{15}{51}\)
\(C=\dfrac{8}{51}\)
3 tháng 4 2016
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(M=\frac{1}{2}\left(1-\frac{1}{51}\right)\)
M=\(\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)
3 tháng 4 2016
\(M=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(M=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(M=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{51}\right)\)
\(M=\frac{1}{2}.\frac{50}{51}\)
\(M=\frac{25}{51}\)
\(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\\ =\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}+\dfrac{1}{51}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)=\dfrac{1}{2}.\dfrac{16}{51}\\ =\dfrac{8}{51}\)