Ta có:\(f\left(x\right)-1=\left(x-1\right)^3\)
\(=>A+\frac{1}{2}=\left(\frac{1}{112}-1\right)^3+\left(\frac{2}{112}-1\right)^3+\left(\frac{3}{112}-1\right)^3+...\left(\frac{111}{112}-1\right)^3\)
\(A+\frac{1}{2}=-\frac{1^3+2^3+3^3+...+111^3}{112^3}=-\frac{\frac{111^2\left(111+1\right)^2}{4}}{112^3}=-\frac{111^2}{4\cdot112}=-\frac{12321}{448}\)
\(A=-\frac{12321}{448}-\frac{1}{2}=-\frac{12545}{448}\)
cho \(a\)và \(1-a\), ta có:
\(f\left(1-a\right)=\frac{\left(1-a\right)^3}{3\left(1-a\right)^2-3\left(1-a\right)+1}=\frac{\left(1-a-1\right)^3}{3-6a+a^2-3+3a+1}+1=1-\frac{a^3}{3a^3-3a+1}=1-f\left(a\right)\)
hay \(f\left(a\right)+f\left(1-a\right)=1\)
\(=>A=f\left(\frac{1}{112}\right)+f\left(\frac{111}{112}\right)+f\left(\frac{2}{112}\right)+f\left(\frac{110}{112}\right)+...+f\left(\frac{55}{112}\right)+f\left(\frac{57}{112}\right)+f\left(\frac{56}{112}\right)-\frac{1}{2}\)
\(=>A=55+f\left(\frac{1}{2}\right)-\frac{1}{2}=55\) vì \(f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=2f\left(\frac{1}{2}\right)=1\)nên \(f\left(\frac{1}{2}\right)=\frac{1}{2}\)
Vậy \(A=55\)
