\(a)\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}\right)=2\cdot\frac{20}{41}\)
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{x+1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\hept{\begin{cases}x+1=40\\x+2=41\end{cases}\Leftrightarrow\hept{\begin{cases}x=40-1\\x=41-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=39\\x=39\end{cases}}}\)
Vậy x=39
\(b)|x+2016|\ge0\forall x;|x+2017|\ge0\forall x\)
\(\Leftrightarrow x+2016+x+2017+2018=3x\)
\(\Leftrightarrow2x+6051=3x\)
\(\Leftrightarrow6051=3x-2x\)
\(\Leftrightarrow6051=x\)
Vậy x=6051
