Ta thấy vế phải là phương trình bậc 2 nên:
\(\Rightarrow f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(1-x\right)=a\left(1-x\right)^2+b\left(1-x\right)+c=ax^2-x\left(2a+b\right)+a-b+c\)
\(\Rightarrow3f\left(x\right)-f\left(1-x\right)=x^2\left(3a-a\right)+x\left\{3b-\left[-\left(2a+b\right)\right]\right\}+3c-\left(a-b+c\right)\)
\(=x^2+1\)
\(\Rightarrow2a.x^2+2x\left(a+2b\right)-a+b-2x=x^2+1\)
\(\Rightarrow\left\{{}\begin{matrix}2a=1\\a+2b=0\\-a+b-2c=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{2}\\b=-\frac{1}{4}\\c=-\frac{7}{8}\end{matrix}\right.\)
\(\Rightarrow f\left(x\right)=\frac{1}{2}x^2-\frac{1}{4}x-\frac{7}{8}\)
Vậy .......................................................................
