x+\(\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
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NP
7
23 tháng 3 2018
theo đề bài ta có:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=1\)
TC
23 tháng 3 2018
đặt A=4/5.9+4/9.13+4/13.17+...+4/41.45
=1/5-1/9+1/9-1/13+1/13-1/17+...+1/41-1/45
=1/5-1/45
=8/45
suy ra x+8/45=-37/45
suy ra x=-1
TT
1
DP
27 tháng 7 2017
\(x+\frac{2}{5.9}+\frac{2}{9.13}+\frac{2}{13.17}+...+\frac{2}{41.45}=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{2}{4}\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{45}\right)\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\left[\frac{1}{2}\cdot\frac{8}{45}\right]=\frac{-37}{45}\)
\(\Leftrightarrow x+\frac{4}{45}=\frac{-37}{45}\)
\(\Leftrightarrow x=-\frac{41}{45}\)
MT
12 tháng 8 2015
\(\frac{x+1}{5}=\frac{-10}{16}\Rightarrow x+1=\frac{5.\left(-10\right)}{16}=\frac{-25}{8}\Rightarrow x=\frac{-25}{8}-1=-\frac{33}{8}\)
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=-1\)
BD
10 tháng 8 2018
\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)
\(\Leftrightarrow x=\frac{7.45}{21}=15\)
17 tháng 3 2021
Ta có : \(\frac{7}{x-2005}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)\)
\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}-\frac{8}{45}=\frac{7}{15}\)
\(\Rightarrow x-2005=15\Rightarrow x=15+2005=2020\)
Vậy x =2020
NH
3
19 tháng 3 2017
x + 4/5.9 + 4/9.13 + ... + 4/41.45 = -37/45
<=> x + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/41 - 1/45= -37/45
<=> x + 1/5 - 1/45 = -37/45
<=> x + 9/45 = -36/45
<=>x= -45/45=-1
6 tháng 3 2016
bài này mà cũng đăng dễ ẹc
x + 4/5.9 + 4/9.13 + 4/13.17 + ... + 4/41.45 =-37/45
x+( 1/5 - 1/9 +1/9 - 1/13 + 1/13 - 1/17 + ... + 1/41 - 1/45 )= -37/45
x+ (1/5 - 1/45 )=-37/45
x+8/45 = -37 / 45
x=-37/45 -8/45
x=-45/45=-1
\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)
\(\Leftrightarrow x=-\frac{43}{45}\)