Tính giá trị biểu thức:
\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
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22 tháng 3 2018
hình như là 32 chứ k f 33
\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)
\(B=\frac{\left(1\cdot1\right)\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)}{\left(1\cdot2\right)\left(2\cdot3\right)\left(3\cdot4\right)\left(4\cdot5\right)}\)
\(B=\frac{\left(1\cdot2\cdot3\cdot4\right)\left(1\cdot2\cdot3\cdot4\right)}{\left(1\cdot2\cdot3\cdot4\right)\left(2\cdot3\cdot4\cdot5\right)}\)
\(=\frac{1}{5}\)
22 tháng 3 2018
\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)
\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5}\)
\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1^2\cdot2^2\cdot3^2\cdot4^2\cdot5}=\frac{1}{5}\)
S
3 tháng 5 2017
\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1}{5}\)
\(N=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
3 tháng 5 2017
N=1/2x(1/3-1/5+1/5-1/7+....+1/99-1/101)
N=1/2x(1/3-1/101)
N=1/2x98/101
N=49/101
3 tháng 5 2016
Ta có:
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}=\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}=\frac{1}{5}\)
4 tháng 2 2017
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
VH
3
DN
19 tháng 12 2016
Ta có \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}\)
\(\Rightarrow A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow A=2\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=2\left(\frac{2016}{2017}\right)\)
\(\Rightarrow A=\frac{4032}{2017}\)
PM
19 tháng 12 2016
Ta có:\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2016\cdot2017}\)
\(=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+....+\frac{2}{2016}-\frac{2}{2017}\)
\(=\frac{2}{1}-\frac{2}{2017}=2-\frac{2}{2017}=\frac{4034}{2017}-\frac{2}{2017}=\frac{4032}{2017}\)
LC
17 tháng 6 2015
Ta có: \(1-\frac{2}{n.\left(n+1\right)}\)
=\(\frac{n.\left(n+1\right)-2}{n\left(n+1\right)}\)
=\(\frac{n^2+n-2}{n.\left(n+1\right)}\)
=\(\frac{\left(n^2-1\right)+\left(n-1\right)}{n.\left(n+1\right)}\)
=\(\frac{\left(n-1\right).\left(n+1\right)+\left(n-1\right)}{n.\left(n+1\right)}\)
=\(\frac{\left(n-1\right).\left(n+1+1\right)}{n.\left(n+1\right)}\)
=\(\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\)
=>\(1-\frac{2}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\left(1\right)\)
Lại có: \(M=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)....\left(1-\frac{2}{99.100}\right)\)
=> \(M=\left(1-\frac{2}{2.\left(2+1\right)}\right).\left(1-\frac{2}{3.\left(3+1\right)}\right).\left(1-\frac{2}{4.\left(4+1\right)}\right)....\left(1-\frac{2}{99.\left(99+1\right)}\right)\left(2\right)\)
Thay (1) vào (2) ta được:
\(M=\frac{\left(2-1\right).\left(2+2\right)}{2.\left(2+1\right)}.\frac{\left(3-1\right).\left(3+2\right)}{3.\left(3+1\right)}.\frac{\left(4-1\right).\left(4+2\right)}{4.\left(4+1\right)}...\frac{\left(99-1\right).\left(99+2\right)}{99.\left(99+1\right)}\)
=> \(M=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{98.101}{99.100}\)
=> \(M=\frac{1.4.2.5.3.6....98.101}{2.3.3.4.4.5....99.100}\)
=> \(M=\frac{\left(1.2.3....98\right).\left(4.5.6....101\right)}{\left(2.3.4....99\right).\left(3.4.5....100\right)}\)
=> \(M=\frac{1.101}{99.3}\)
=> \(M=\frac{101}{297}\)
Vậy \(M=\frac{101}{297}\)
2 tháng 5 2016
A = \(\frac{5}{1.2}\) + \(\frac{5}{2.3}\) +........+\(\frac{5}{99.100}\)
A = 5.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +......+\(\frac{1}{99.100}\) )
A = 5. ( \(\frac{1}{1}\) - \(\frac{1}{2}\) +\(\frac{1}{2}-\frac{1}{3}\) +......+\(\frac{1}{99}-\frac{1}{100}\) )
A= 5. (\(1-\frac{1}{100}\))
A= 5.\(\frac{99}{100}\)
A= \(\frac{99}{20}\)
23 tháng 3 2017
B = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............+ \(\frac{1}{9.10}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...................+\(\frac{1}{9}\)- \(\frac{1}{10}\)
= \(\frac{1}{2}\) - \(\frac{1}{10}\)
= \(\frac{2}{5}\)
3 tháng 2 2017
\(A-1=\frac{1}{1.2}+\frac{1}{2.3}..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}\)\(=\frac{99}{100}\)
\(A=1+\frac{99}{100}=\frac{199}{100}\)
CC
2
PN
9 tháng 5 2016
Đặt biểu thức trên là A
Ta có: A =(1^2 . 2^2 . 3^2 . 4^2)/(1.2.2.3.3.4.4.5)
= [(1.2.3.4).(1.2.3.4)] / [(1.2.3.4).(2.3.4.5)]
= 1/5
Vậy A = 1/5
9 tháng 5 2016
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)
=\(\frac{1}{5}\)
sorry mình nhầm
ta có:
M=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).\(\frac{4^2}{4.5}\)
=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)
=\(\frac{1}{5}\)
vậy M=\(\frac{1}{5}\)
\(M=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)