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23 tháng 3 2018

theo đề bài ta có:

\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)

\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)

\(x+\frac{8}{45}=\frac{-37}{45}\)

\(x=\frac{-37}{45}-\frac{8}{45}\)

\(x=\frac{-45}{45}=1\)

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

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

19 tháng 3 2017

CÓ: x+1/5-1/9+1/9-1/13+....+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

vậy x=-1

6 tháng 6 2019

\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)

\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)

\(\Leftrightarrow x=-\frac{43}{45}\)

10 tháng 8 2018

\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)

\(\Leftrightarrow x=\frac{7.45}{21}=15\)

7 tháng 6 2020

\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{9}{45}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)

\(\frac{7}{x}=\frac{21}{45}\)

\(\frac{7}{x}=\frac{7}{15}\)

\(\Rightarrow x=15\)

Vậy \(x=15\).

7 tháng 6 2020

\(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)

=> \(\frac{7}{x}+4\left(\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+....+\frac{1}{41\cdot45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+4\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+4\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+4\cdot\frac{32}{45}=\frac{29}{45}\)

=> \(\frac{7}{x}+\frac{128}{45}=\frac{29}{45}\)

=> \(\frac{7}{x}=-\frac{11}{5}\)

Đến đây tự giải quyết :))

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

17 tháng 3 2021

sry =29/45 nha

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 

16 tháng 3 2019

a) \(x+\)\(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)

\(\Rightarrow x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}=-\frac{4}{5}\)

\(\Rightarrow x=\frac{-3}{5}\)

b) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\)

\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(\Rightarrow2A=1-\frac{1}{2005}\)

\(\Rightarrow2A=\frac{2004}{2005}\)

\(\Rightarrow A=\frac{1002}{2005}\)

16 tháng 3 2019

Tính tổng:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\) 

\(\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003+2005}\right)\)  

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{2003}-\frac{1}{2005}\right)\) 

\(\frac{1}{2}\left(1-\frac{1}{2005}\right)\)

\(\frac{1}{2}\cdot\frac{2004}{2005}\)  

\(\frac{1002}{2005}\) 

k nha