Question

Tìm x biết :

\(\frac{7}{x-2005}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\)với x khác 2005

Answer 1

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

Answer 2

sry =29/45 nha

Answer 3

cái j bằng 29/ 45 cơ

Answer 4

qetryuioplkjhgfhjjkklmnbvzươ-=ơ/.,`134567890