KT
5
K
Khách
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PK
3
31 tháng 7 2016
\(K=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)
\(=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\right):2\)
= \(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right):2\)
= \(\left(1-\frac{1}{49}\right):2\)
\(=\frac{48}{49}:2\) \(\frac{24}{49}\)
TT
2
20 tháng 8 2016
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\Rightarrow x=\frac{49}{24}\)
20 tháng 8 2016
\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{1}{x}\)
\(\frac{24}{49}=\frac{1}{x}\)\(\Rightarrow x=\frac{49}{24}\)
9 tháng 4 2018
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
9 tháng 4 2018
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
NQ
0
MC
4 tháng 5 2016
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
D
6
EC
8 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\frac{98}{99}\)
\(=\frac{98}{297}\)
Chuc bn học tốt
8 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
TC
2
10 tháng 4 2018
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)
=\(1-\frac{1}{2011}\)
=\(\frac{2010}{2011}\)
10 tháng 4 2018
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\)
\(=\frac{1\cdot2}{2\cdot1\cdot3}+\frac{1\cdot2}{2\cdot3\cdot5}+\frac{1\cdot2}{2\cdot5\cdot7}+...+\frac{1\cdot2}{2\cdot2009\cdot2011}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2009\cdot2011}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{2011}\right)\)= .......
Mình không chắc là đúng đâu nha
23 tháng 4 2017
= 2 x [1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +1/7 -1/9 + .., +1/99 - 1/101
= 2 x [ 1 - 1/101 ]
= 2 x 100/101
= 200/101
t cho mik nha
23 tháng 4 2017
\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+.........+\(\frac{2}{99.101}\)
=\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+....+\(\frac{1}{99}\)-\(\frac{1}{101}\)
= 1 - \(\frac{1}{101}\)= \(\frac{100}{101}\)
A
5
KN
24 tháng 2 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Leftrightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(\Leftrightarrow2A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}\)
\(\Leftrightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(\Leftrightarrow2A=1-\frac{1}{99}\)
\(\Leftrightarrow2A=\frac{99}{99}-\frac{1}{99}\)
\(\Leftrightarrow2A=\frac{98}{99}\)
\(\Leftrightarrow A=\frac{98}{99}\div2\)
\(\Leftrightarrow A=\frac{49}{99}\)
24 tháng 2 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97+99}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)
\(A=\left(1-\frac{1}{99}\right)+\left(-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{95}-\frac{1}{97}\right)\)
\(A=\frac{98}{99}+0\)
\(A=\frac{98}{99}\)
\(K=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{49}\right)\)
\(=\frac{1}{2}.\frac{48}{49}\)
\(=\frac{24}{49}\)
\(K\times2=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\)
\(K\times2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\)
\(K\times2=\frac{48}{49}\)
\(K=\frac{48}{49}\div2=\frac{24}{49}\)