M=\(\frac{16}{1.5}+\frac{16}{5.9}+....+\frac{16}{2017.2021}\) ; N =\(\frac{1}{1.7}+\frac{1}{7.13}+....+\frac{1}{2007.2013}\)
K
Khách
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Những câu hỏi liên quan
20 tháng 4 2017
\(\frac{4}{3.5}+\frac{8}{5.9}+\frac{12}{9.15}+...+\frac{32}{n\left(n+16\right)}=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{5}\right)+2\left(\frac{1}{5}-\frac{1}{9}\right)+2\left(\frac{1}{9}-\frac{1}{15}\right)+...+2\left(\frac{1}{n}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(\frac{1}{3}-\frac{1}{n+16}=\frac{8}{25}\)
\(\frac{1}{n+16}=\frac{1}{75}\)
\(\Rightarrow n+16=75\)
\(\Rightarrow n=59\)
9 tháng 1 2018
\(=\frac{4^5.\left(3^2\right)^4}{\left(4^2\right)^2.\left(3^3\right)^3}=\frac{4^5.3^8}{4^4.3^9}=\frac{4}{3}\)
9 tháng 1 2018
\(=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^4\right)^2\cdot\left(3^3\right)^3}\)
\(=\frac{2^{10}\cdot3^8}{2^8\cdot3^9}\)
\(=\frac{4}{3}\)
MT
12 tháng 8 2015
\(\frac{x+1}{5}=\frac{-10}{16}\Rightarrow x+1=\frac{5.\left(-10\right)}{16}=\frac{-25}{8}\Rightarrow x=\frac{-25}{8}-1=-\frac{33}{8}\)
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=-1\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2019
\(cos\left(2\pi+\frac{\pi}{16}\right).sin\frac{5\pi}{16}.cos\frac{5\pi}{16}.cos\left(\frac{\pi}{2}-\frac{\pi}{16}\right)\)
\(=\frac{1}{4}.2cos\frac{\pi}{16}.sin\frac{\pi}{16}.2sin\frac{5\pi}{16}.cos\frac{5\pi}{16}\)
\(=\frac{1}{4}sin\frac{2\pi}{16}.sin\frac{10\pi}{16}=\frac{1}{4}sin\frac{\pi}{8}.sin\frac{5\pi}{8}\)
\(=\frac{1}{4}sin\frac{\pi}{8}.sin\left(\frac{\pi}{2}+\frac{\pi}{8}\right)\)
\(=\frac{1}{4}sin\frac{\pi}{8}.cos\frac{\pi}{8}=\frac{1}{8}sin\frac{2\pi}{8}\)
\(=\frac{1}{8}sin\frac{\pi}{4}=\frac{\sqrt{2}}{16}\)
Đề sai hoặc bạn gõ thiếu số 1 ở dưới mẫu
31 tháng 5 2015
\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)
\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)
15 tháng 8 2016
A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
3 tháng 2 2022
a: \(=\dfrac{2^4\cdot3^6\cdot2\cdot3}{2^4\cdot3^6}=6\)
b: \(=\dfrac{2^{20}\cdot3^{20}}{2^{18}\cdot3^{18}}=2^2\cdot3^2=36\)
c: \(=\dfrac{12^5\cdot13}{12^6\cdot13}-\dfrac{12^8\cdot\left(-11\right)}{12^9\cdot\left(-11\right)}=\dfrac{1}{12}-\dfrac{1}{12}=0\)
CY
7
2 tháng 4 2019
\(A=4\left(\frac{1}{20}-\frac{1}{80}\right)=4.\frac{3}{80}=60\)
2 tháng 4 2019
\(A=\frac{16}{20\cdot24}+\frac{16}{24\cdot28}+\frac{16}{28\cdot32}+...+\frac{16}{76\cdot80}\)
\(A=4\left[\frac{4}{20\cdot24}+\frac{4}{24\cdot28}+\frac{4}{28\cdot32}+...+\frac{4}{76\cdot80}\right]\)
\(A=4\left[\frac{1}{20}-\frac{1}{24}+...+\frac{1}{76}-\frac{1}{80}\right]\)
\(A=4\left[\frac{1}{20}-\frac{1}{80}\right]\)
\(A=4\left[\frac{4}{80}-\frac{1}{80}\right]=4\cdot\frac{3}{80}=\frac{4\cdot3}{80}=\frac{1\cdot3}{20}=\frac{3}{20}\)
\(M=\frac{16}{1.5}+\frac{16}{5.9}+........+\frac{16}{2017.2021}\)
\(M=4.\left(\frac{4}{1.5}+\frac{4}{5.9}+.......+\frac{4}{2017.2021}\right)\)
\(M=4.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.........+\frac{1}{2017}-\frac{1}{2021}\right)\)
\(M=4.\left(1-\frac{1}{2021}\right)\)
\(M=4.\frac{2020}{2021}\)
\(M=\frac{8080}{2021}\)
\(N=\frac{1}{1.7}+\frac{1}{7.13}+.......+\frac{1}{2007.2013}\)
\(N=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+........+\frac{6}{2007.2013}\right)\)
\(N=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+......+\frac{1}{2007}-\frac{1}{2013}\right)\)
\(N=\frac{1}{6}.\left(1-\frac{1}{2013}\right)\)
\(N=\frac{1}{6}.\frac{2012}{2013}\)
\(N=\frac{1006}{6039}\)
\(N=\frac{1}{1.7}+\frac{1}{7.13}+...+\frac{1}{2007.2013}\)
\(N=\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{2007}-\frac{1}{2013}\)
\(N=1-\frac{1}{2013}\)
\(N=\frac{2012}{2013}\)