\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{x}-\frac{2}{\left(x+2\right)}=\frac{2015}{2016}\)
\(\Rightarrow2-\frac{2}{x+2}=\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{x+2}=2-\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{x+2}=\frac{2017}{2016}\)
\(\Rightarrow2017.\left(x+2\right)=2.2016\)
\(\Rightarrow2017x+4034=4032\)
\(\Rightarrow2017x=-2\)
\(\Rightarrow x=-\frac{2}{2017}\)
Vậy......
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{x\cdot\left(x+2\right)}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(=1-\frac{1}{x+2}=\frac{2015}{2016}\)
=>\(\frac{1}{x+2}=\frac{1}{2016}\)
=>\(x+2=2016\)
=>\(x=2014\)
Vậy.......
Ta có:
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2015}{2016}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(\Rightarrow x+2=1\div\frac{1}{2016}=2016\)
\(\Rightarrow x=2016-2=2014\)
Vậy giá trị của x là 2014.
