\(B=\frac{x^2+x+1}{x}=\frac{7}{2}\)
\(\frac{x^2+x+1}{x}-\frac{7}{2}=0\)
\(\frac{x^2+x+1}{x}-\frac{3,5x}{x}=0\)
\(\frac{x^2-2,5x+1}{x}=0\)<=> \(x^2-2,5x+1=0\)
\(\left(\frac{x}{2}-1\right)\left(2x-1\right)=0\)
Th1: x/2 -1 = 0 <= > x = 2
TH2: 2x - 1 = 0 < = > x = 1/2
Vậy x thuộc {1/2 ; 2}