thử làm:))
\(\hept{\begin{cases}f\left(x\right)+g\left(x\right)=5x^2-2x+3\\f\left(x\right)-g\left(x\right)=x^2-2x+5\end{cases}}\)
\(\Rightarrow f\left(x\right)+g\left(x\right)+f\left(x\right)-g\left(x\right)=\left(5x^2-2x+3\right)+\left(x^2-2x+5\right)\)
\(\Rightarrow2\cdot f\left(x\right)=6x^2-4x+8\)
\(\Rightarrow f\left(x\right)=3x^2-2x+4\)
\(\Rightarrow\hept{\begin{cases}3x^2-2x+4+g\left(x\right)=5x^2-2x+3\\3x^2-2x+4-g\left(x\right)=x^2-2x+5\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}g\left(x\right)=2x^2-1\\g\left(x\right)=2x^2-1\end{cases}}\)
Vậy ...