Question

Tìm x biết \(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+............+\(\frac{1}{x.\left(x+2\right)}\)ĐS:\(\frac{1005}{2011}\)mình đang cần gấp nếu có bạn nào làm được thì làm nhanh giúp mình nha

Answer 1

.........................

= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... +  \(\frac{2}{x.\left(x+2\right)}\) )

= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) +  \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)-  \(\frac{1}{x+2}\) ) 

= ................ 

Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)

Answer 2

 Các bạn ơi! giải chi tiết ra cho mình luôn nha 

Answer 3

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{1005}{2011}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{1005}{2011}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1006}{2011}\)

\(\Rightarrow x+2=2011\Rightarrow x=2011-2\Rightarrow x=2009\)

Answer 4

Gọi biểu thức là A

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}=\frac{1005}{2011}.2\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2010}{2011}\)

\(2A=\frac{1}{1}-\frac{1}{x+2}=\frac{2010}{2011}\)

\(A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{x+2}\right)=\frac{2010}{2011}:2=\frac{1005}{2011}\)

\(A=\frac{1}{2}.\frac{1}{1}-\frac{1}{2}.\frac{1}{x+2}=\frac{1005}{2011}\)

\(A=\frac{1}{2}-\frac{1}{2x+4}=\frac{1005}{2011}\)

\(\frac{1}{2x+4}=\frac{1}{2}-\frac{1005}{2011}=\frac{2011}{4022}-\frac{2010}{4022}=\frac{1}{4002}\)

\(\Rightarrow2x+4=4002\)

\(2x=4002-4=3098\)

\(x=3098:2=1549\)