Đáp án:
\(x\notin\mathbb{R}\)
Giải thích các bước giải:
\(\dfrac{2x^2}{3-\sqrt{(9+2x)^2}}=x+9 \qquad (x\ne -3, x\ne -6)\\\Leftrightarrow \dfrac{2x^2}{3-|9+2x|}=x+9\\\Leftrightarrow 2x^2 =(x+9)(3-|9+2x|)\\\Leftrightarrow 2x^2 - (3x-x\times|9+2x|+27-9\times|9+2x|)=0\\\Leftrightarrow 2x^2 - 3x+x\times|9+2x|-27+9\times|9+2x|=0\\\Leftrightarrow \left[ \begin{array}{l}2x^2- 3x + x(9+2x)-27+9(9+2x)=0,\quad 9+2x\ge 0\\\\2x^2 -3x + x[-(9+2x)]-27 +9[-(9+2x)]=0,\quad 9+2x<0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}2x^2 +12x +27=0,\quad 2x\ge -9\\\\-30x =108,\quad 2x<-9\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{-12\pm\sqrt{12^2 -4\times 2\times 2\times 27}}{2\times 2}=\dfrac{-12\pm\sqrt{-72}}{4}\quad(!)\quad x\ge -\dfrac{9}{2}\\x=-\dfrac{18}{5},\quad x<-\dfrac{9}{2}\end{array} \right.\ \\\Leftrightarrow \left[ \begin{array}{l}x\notin\mathbb{R}\\x\in \emptyset\end{array} \right.\\\Leftrightarrow x\notin\mathbb{R}\)
