Đặt \(f\left(x\right)=ax+b\Rightarrow\left\{{}\begin{matrix}f\left(2x-1\right)=a\left(2x-1\right)+b=2ax-a+b\\f\left(2x+1\right)=a\left(2x+1\right)+b=2ax+a+b\end{matrix}\right.\)
\(f\left(2x-1\right)+f\left(2x+1\right)-f\left(x\right)=x+3\)
\(\Leftrightarrow2ax-a+b+2ax+a+b-ax-b=x+3\)
\(\Leftrightarrow3ax-x+b-3=0\)
\(\Leftrightarrow\left(3a-1\right)x+\left(b-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a-1=0\\b-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{3}\\b=3\end{matrix}\right.\) \(\Rightarrow f\left(x\right)=\frac{1}{3}x+3\)