(2x+1)²+(2-x)(2x-1)=0
⇔4x²+4x+1+4x-2-2x²+x=0
⇔2x²+9x-1=0
⇔2(x²+$\frac{9}{2}$x-$\frac{1}{2}$)=0
⇔2[(x+$\frac{9}{4}$)²-$\frac{89}{16}$]=0
⇔2(x+$\frac{9}{4}$+$\frac{√89}{4}$)(x+$\frac{9}{4}$-$\frac{√89}{4}$)=0
⇔2(x+$\frac{9+√89}{4}$)(x+$\frac{9-√89}{4}$)=0
⇔\(\left[ \begin{array}{l}x+\frac{9+√89}{4}=0\\x+\frac{9-√89}{4}=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=-\frac{9+√89}{4}\\x=-\frac{9-√89}{4}\end{array} \right.\)
Vậy S={-$\frac{9+√89}{4}$;-$\frac{9-√89}{4}$}.
