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\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(M=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99})\)
\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(M=2.\dfrac{32}{99}\)
\(M=\dfrac{64}{99}\)
http://vietjack.com/giai-sach-bai-tap-toan-6/bai-95-trang-28-sach-bai-tap-toan-6-tap-2.jsp
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+.....+\dfrac{2}{97.99}\)
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 = \(\dfrac{98}{99}\)
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 = \(\dfrac{32}{99}\)
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
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{39}\)
tính
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{39}\)
\(=(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13})+\dfrac{2}{39}\)
\(=(\dfrac{1}{3}-\dfrac{1}{13})+\dfrac{2}{39}\)
\(=\dfrac{10}{39}+\dfrac{2}{39}\)
\(=\dfrac{4}{13}\)
gọi biểu thức đó là A
A=\(1.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)+\dfrac{2}{39}\)
A= \(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)+\dfrac{2}{39}=\dfrac{4}{13}\)
mk nhanh nhất nha bạn
\(S=\dfrac{5-3}{5.3}+\dfrac{7-5}{7.5}....+\dfrac{25-23}{23.25}\)
\(S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{23}-\dfrac{1}{25}\)
\(S=\dfrac{1}{3}-\dfrac{1}{25}=\dfrac{25-3}{3.25}=\dfrac{7}{25}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{13}-\dfrac{1}{15}\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)
Chúc bạn học tốt!!!
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)
= \(\dfrac{1}{3}-\dfrac{1}{15}\)
= \(\dfrac{4}{15}\)
\(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\left(1-\dfrac{1}{99}\right)-x=-\dfrac{100}{99}\)
\(\dfrac{98}{99}-x=-\dfrac{100}{99}\)
\(x=\dfrac{98}{99}-\left(-\dfrac{100}{99}\right)\)
\(x=\dfrac{198}{99}\)
Vậy \(x=\dfrac{198}{99}\)
\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{33}{99}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{95}-\dfrac{2}{97}+\dfrac{2}{97}-\dfrac{2}{99}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{2}{99}=\dfrac{64}{99}\)
chúc bạn học tốt
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 ...
Thanks