`~rai~`
\(A=\{x\in\mathbb{R}|2x^2+x+4=4x^2-4x+1\}\\\text{Ta có:}2x^2+x+4=4x^2-4x+1\\\Leftrightarrow 4x^2-4x+1-2x^2-x-4=0\\\Leftrightarrow 2x^2-5x-3=0\\\Leftrightarrow (x-3)(2x+1)=0\\\Leftrightarrow \left[\begin{array}{I}x-3=0\\2x+1=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=3\\x=-\dfrac{1}{2}\end{array}\right.\quad\text{(thỏa mãn)}\\\Rightarrow A=\left\{-\dfrac{1}{2};3\right\}\\\Rightarrow \text{Tập hợp A có 2 phần tử.}\)
