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\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(M=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99})\)
\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(M=2.\dfrac{32}{99}\)
\(M=\dfrac{64}{99}\)
http://vietjack.com/giai-sach-bai-tap-toan-6/bai-95-trang-28-sach-bai-tap-toan-6-tap-2.jsp
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
Giải:
Biến đổi vế trái BĐT:
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
Vì \(\dfrac{32}{99}>\dfrac{32}{100}\)
\(\Leftrightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}>\dfrac{32}{100}\)
\(\Leftrightarrow\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}>32\%\)
Vậy ...
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+.....+\dfrac{2}{97.99}\)
Ta thấy:\(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{1.3
}\)
\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{3.5}\)
............\(\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{2}{97.99}\)
\(\Rightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{97}-\dfrac{1}{99}\) =\(\dfrac{1}{1}-\dfrac{1}{99}\)
=\(\dfrac{99}{99}-\dfrac{1}{99}\)
=\(\dfrac{98}{99}\)
Vậy A=\(\dfrac{98}{99}\)
A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
A = \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(1-\dfrac{1}{99}\)
A = \(\dfrac{98}{99}\)
\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)
\(\dfrac{2}{1\cdot3}=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
\(\dfrac{2}{5\cdot7}=\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{7}{35}-\dfrac{5}{35}=\dfrac{2}{35}\)
và cứ như thế đến số cuối
2)
a) M = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{99}\)
M = \(\dfrac{32}{99}\)
1) Quãng đường AB là :
\(36\dfrac{1}{4}\)km/h . 3.2h = 116 ( km )
Thời gian người ấy đi từ A đến B lúc về là :
116 : 40 = 2.9 ( giờ )
Đ/S : 2.9 giờ
`2/(3.5)+2/(5.7)+....+2/(2015.2017)`
`=1/3-1/5+1/5-1/7+....+1/2016-1/2017`
`=1/3-1/2017=2014/6051`
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{2017}\)
\(=\dfrac{2017}{6051}-\dfrac{3}{6051}=\dfrac{2014}{6051}\)
Ta có
M = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
M = \(\dfrac{1}{3}-\dfrac{1}{99}\)
M = \(\dfrac{32}{99}\)
Vậy M = \(\dfrac{32}{99}\)
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(M=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(M=\dfrac{1}{3}-\dfrac{1}{99}\)
\(M=\dfrac{32}{99}\)