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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: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{13}-\dfrac{1}{15}\)
(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!!!
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)
= \(\dfrac{1}{3}-\dfrac{1}{15}\)
= \(\dfrac{4}{15}\)
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
1. Ta có: \(\left|x\right|=7\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
Vậy \(x\in\left\{\pm7\right\}\)
2. \(M=\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\left(\dfrac{13}{39}-\dfrac{3}{39}\right)\)
\(\Rightarrow M=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{1.10}{2.39}=\dfrac{5}{39}\)
Tick mk vs! Thank nhiều!
1. Theo đb ta có: |x|=7
=> Có 2 TH:\(\left\{{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\) \(\in Z\)
Vậy x=7 \(\veebar\) x= -7 ( x\(\in\) Z) thì |x|=7
2. \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
Đặt A= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
Ta thấy: \(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{3.5}\)
\(\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{2}{5.7}\)
... \(\dfrac{1}{11}-\dfrac{1}{13}=\dfrac{2}{11.13}\)
=> 2D=2(\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\))
<=> 2D= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
<=>2D=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
<=> 2D= \(\dfrac{1}{3}-\dfrac{1}{13}\)
<=>2D= \(\dfrac{13}{39}-\dfrac{3}{39}\)
<=>2D=\(\dfrac{10}{39}\)
=> D= \(\dfrac{10}{39}:2\)
<=> D= \(\dfrac{10}{39}.\dfrac{1}{2}\)
<=> D=\(\dfrac{5}{39}\)
Vậy D= \(\dfrac{5}{39}\)
_ Chc bn hk tốt_
\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.100}\)
\(B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99.100}+\dfrac{1}{99.100}\)
\(B=\dfrac{1}{3}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B=\dfrac{1}{3}-\dfrac{2}{100}+\dfrac{1}{99}\)
\(B=\dfrac{1}{3}-\dfrac{1}{50}+\dfrac{1}{99}\)
Đến đây thì hết tính hợp lý được rồi:v
\(B=\dfrac{34}{99}-\dfrac{1}{50}\)
\(B=\dfrac{1601}{4950}\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2020.2022}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}\)
\(=\dfrac{2021}{2022}\)
`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`
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{2017}\)
\(=\dfrac{2017}{6051}-\dfrac{3}{6051}=\dfrac{2014}{6051}\)
1: \(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=1/2*10/39
=5/39
2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)
\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(M=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99})\)
\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(M=2.\dfrac{32}{99}\)
\(M=\dfrac{64}{99}\)
http://vietjack.com/giai-sach-bai-tap-toan-6/bai-95-trang-28-sach-bai-tap-toan-6-tap-2.jsp
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{39}\)
\(=(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13})+\dfrac{2}{39}\)
\(=(\dfrac{1}{3}-\dfrac{1}{13})+\dfrac{2}{39}\)
\(=\dfrac{10}{39}+\dfrac{2}{39}\)
\(=\dfrac{4}{13}\)
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