2 câu trả lời
Ta có: \((2x+1)^3-(2x-3)(4x^2+6x+9)=34\) \(\Leftrightarrow 8x^3+12x^2+6x+1-8x^3-12x^2-18x+12x^2+18x+27=34\) \(\Leftrightarrow 12x^2+6x-6=0\) \(\Leftrightarrow 12x^2+12x-6x-6=0\) \(\Leftrightarrow 12x(x+1)-6(x-1)=0\) \(\Leftrightarrow (x+1)(12x-6)=0\) \(\Leftrightarrow\left\{ \begin{array}{l} x+1=0 \\ 12x-6=0 \end{array} \right .\Leftrightarrow\left\{ \begin{array}{l} x=-1 \\ x=\dfrac{1}{2} \end{array} \right .\)
$(2x+1)^3-(2x-3)(4x^2+6x+9)=34\\\to8x^3+12x^2+6x+1-(8x^3-27)=34\\\to8x^3+12x^2+6x+1-8x^3+27-34=0\\\to 12x^2+6x-6=0\\\to(12x^2+12x)+(-6x-6)=0\\\to12x(x+1)-6(x+1)=0\\\to(x+1)(12x-6)=0\\\to\left[ \begin{array}{l}x+1=0\\12x-6=0\end{array} \right.\\\to\left[ \begin{array}{l}x=-1\\x=\cfrac{1}{2}\end{array} \right.\\\to S=\bigg\{-1;\cfrac{1}{2}\bigg\}$