Xét hàm \(g\left(x\right)=f\left(x\right)-10x\)
\(\Rightarrow g\left(1\right)=f\left(1\right)-10.1=10-10=0\)
Tương tự \(g\left(2\right)=0\) ; \(g\left(3\right)=0\)
\(\Rightarrow g\left(x\right)\) luôn có 3 nghiệm \(x=\left\{1;2;3\right\}\)
\(\Rightarrow g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-a\right)\) với a là số thực bất kì
\(\Rightarrow f\left(x\right)=g\left(x\right)+10x=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-a\right)+10x\)
\(\Rightarrow f\left(12\right)=990\left(12-a\right)+120=12000-990a\)
\(f\left(-8\right)=-990\left(-8-a\right)-80=7840+990a\)
\(\Rightarrow\frac{f\left(12\right)+f\left(-8\right)}{10}+15=\frac{12000-990a+7840+990a}{10}+15=1999\)
