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Ta có:\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{340}=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)
= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{17}-\frac{1}{20}\right)=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
\(\Rightarrow\left[\frac{1}{2\times5}+\frac{1}{5\times8}+...+\frac{1}{17\times20}\right]\cdot\frac{2x}{10}\)
\(\Rightarrow\left[\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\right]\cdot20=\frac{2x}{10}\)
\(\Rightarrow\left[\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]\right]\cdot20=\frac{2x}{10}\)
\(\Rightarrow\left[\frac{1}{3}\cdot\frac{9}{20}\right]\cdot20=\frac{2x}{10}\)
\(\Rightarrow\frac{3}{20}\cdot20=\frac{2x}{10}\)
\(\Rightarrow3\cdot20=\frac{2x}{10}\Leftrightarrow60=\frac{2x}{10}\)
=> 2x = 60*10
=> 2x = 600
=> x = 300
\(\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{340}\right).20=\frac{2x}{10}\)
\(\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right).20=\frac{2x}{10}\)
\(\left[3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\right].20=\frac{2x}{10}\)
\(\left[3.\left(\frac{1}{2}-\frac{1}{20}\right)\right].20=\frac{2x}{10}\)
\(\left(3.\frac{9}{20}\right).20=\frac{2x}{10}\)
\(\frac{27}{20}.20=2x\div10\)
\(27=2x\div10\)
\(x=27\times10\div2\)
\(\Rightarrow x=135\)
\(\frac{3}{4\times7}+\frac{1}{7\times8}+\frac{5}{8\times13}+\frac{2}{13\times15}+\frac{9}{15\times24}\)
= \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{24}\)
= \(\frac{1}{4}-\frac{1}{24}\)
= \(\frac{6}{24}-\frac{1}{24}\)
= \(\frac{5}{24}\)
\(\frac{3}{4.7}+\frac{1}{1.8}+\frac{5}{8.13}+\frac{2}{13.15}+\frac{9}{15.24}\)
Đặt A = ( 3 + 1 + 5 + 2 + 9 ) . \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{24}\)
A = 20 . \(\frac{1}{4}-\frac{1}{24}\)
A = 20 . \(\frac{6}{24}-\frac{1}{24}\)
A = 20 . \(\frac{5}{24}\)
A = \(\frac{100}{24}\)
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# DanLinh
mình cho bạn đó bạn đồng ý nhận lời mời kết bạn từ mình nha!!!!
B1 Nhân biểu tthức trên với 1/2 được
1/6+1/12/+1/20+....+1/110=1/2×3+1/3×4+.....+1/10×11
=1/2-1/3+1/3-1/4+.....+1/10-1/11=1/2-1/11=9/22
B2
B.1/1×2+1/2×3+.....+1/99×100=1-1/2 +1/2-1/3+....+1/99-1/100=1-1/100=99/100
Phần a sai đề phải llà1/2×4+1/4×6+.....+1/8×10 mới làm đc nhé
\(\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+......+\frac{1}{2005.2010}\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+.......+\frac{5}{2005.2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+......+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\left(\frac{1}{5}-\frac{1}{2010}\right)\)
\(=\frac{1}{5}+\frac{1}{5}\frac{401}{2010}\)
\(=\frac{1}{5}+\frac{401}{10050}=\frac{2411}{10050}\)
N = (1/1 - 1/5 + 1/5 -1/10 + ... + 1/2005 - 1/2010 ) x 5
N = (1/1 - 1/2010 ) x5
N = 2009/2010 x5
N = 2009/402
\(ĐặtA=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
Ta có :
\(N=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(N=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)
\(3N=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(3N=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(3N=\frac{1}{2}-\frac{1}{20}\)
\(3N=\frac{9}{20}\)
\(N=\frac{9}{20}:3\)
\(N=\frac{3}{20}\)
Vậy \(N=\frac{3}{20}\)
Chúc bạn học tốt ~
\(N=\frac{1}{10}+\frac{1}{40}+...+\frac{1}{238}+\frac{1}{340}\)
\(N=\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{14.17}+\frac{1}{17.20}\)
\(N=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(N=\frac{1}{2}-\frac{1}{20}\)
\(N=\frac{10}{20}-\frac{1}{20}\)
\(N=\frac{9}{20}\)