Question

tìm x

1/3+1/15+1/35+......+1/x *(x +2)=1005/2011

Answer 1

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{x\times\left(x+2\right)}=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+.....+\frac{1}{x\times\left(x+2\right)}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow\frac{1}{2}\times\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:\frac{1}{2}=\frac{2010}{2011}\)

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

\(\Leftrightarrow x+2=2011\)

\(\Leftrightarrow x=2009\)

Vậy x = 2009

Answer 2

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)

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

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

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

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

\(\Rightarrow x+1=2010\Leftrightarrow x=2009\)

Answer 3

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.......+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\) 

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

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

<=>\(\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

<=> \(1-\frac{1}{x+2}=\frac{1005}{2011}:\frac{1}{2}\)

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

                                                     <=> \(\frac{1}{x+2}=\frac{1}{2011}\) <=>  \(x+2=2011\)<=> \(x=2009\)

vậy \(x=2009\)