Tìm x biết: \(\frac{1}{5X8}+\frac{1}{8X11}+\frac{1}{11X14}+.....+\frac{1}{xX\left(x+3\right)}=\frac{101}{1540}\)
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AN
2
13 tháng 8 2015
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (dấu . là nhân nhé)
=> \(\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)
=> \(\left(\frac{1}{5}-\frac{1}{x+3}\right):3=\frac{101}{1540}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=> \(x+3=308\Rightarrow x=308-3=305\)
ML
13 tháng 8 2015
\(\frac{1}{5x8}+\frac{1}{8x11}+...+\frac{1}{Xx\left(X+3\right)}=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\)
<=>\(\frac{x-2}{5x+15}=\frac{101}{1540}\)
<=>1540x-3080=505x+1515
<=>1035x=4595
<=>x=919/207
NG
2
27 tháng 1 2016
,tao ko biet bai lay giai nhu the nao hay la may cho bai linh tinh hahahahahahahahahahahaaaahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
M
6 tháng 10 2017
\(P=\frac{1}{5x8}+\frac{1}{8x11}+.....+\frac{1}{602x605}\)
\(\Rightarrow3P=\frac{3}{5x8}+\frac{3}{8x11}+......+\frac{3}{602x605}\)
\(\Rightarrow3P=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-.....+\frac{1}{602}-\frac{1}{605}\)
\(\Rightarrow3P=\frac{1}{5}-\frac{1}{605}\)
\(\Rightarrow3P=\frac{24}{121}\)
\(\Rightarrow P=\frac{24}{121}:3\)
\(\Rightarrow P=\frac{8}{121}\)
26 tháng 5 2018
\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}\)
\(=\frac{1}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}+\frac{3}{17\times20}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\frac{9}{20}\)
\(=\frac{3}{20}\)
_Chúc bạn học tốt_
26 tháng 5 2018
Đặt \(A=\frac{1}{2x5}+\frac{1}{5x8}+..+\frac{1}{17x20}\)
\(3xA=3x\left(\frac{1}{2x5}+\frac{1}{5x8}+...+\frac{1}{17x20}\right)\)
\(3xA=\frac{3}{2x5}+\frac{3}{5x8}+....+\frac{3}{17x20}\)
\(3xA=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{17}-\frac{1}{20}\)
\(3xA=\frac{1}{2}-\frac{1}{20}\)
\(3xA=\frac{9}{20}\)
\(\Rightarrow A=\frac{3}{20}\)
16 tháng 6 2016
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow3\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
16 tháng 6 2016
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (x khác 0; khác -3)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{308}\)
=>x+3=308
<=>x=305 (nhận)
Vậy x=305
KT
13 tháng 1 2018
Mk sửa lại đề nha:
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{x\left(x-2\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x-2}-\frac{1}{x}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\)\(\frac{1}{3}-\frac{1}{x}=\frac{101}{1540}:\frac{1}{2}=\frac{101}{770}\)
\(\Leftrightarrow\)\(\frac{1}{x}=\frac{1}{3}-\frac{101}{770}=\frac{467}{2310}\)
\(\Leftrightarrow\)\(x=\frac{2310}{467}\)
P/S: Tham khảo nhé!!!
25 tháng 2 2019
a)Ta có \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
=)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)
toán 6 hả
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