\(\sqrt {2{x^2} - 4x + 9} = 2x - 3\)
Giải chi tiết:
\(\sqrt {2{x^2} - 4x + 9} = 2x - 3\)
\( \Leftrightarrow \left\{ \begin{array}{l}2x - 3 \ge 0\\2{x^2} - 4x + 9 = {\left( {2x - 3} \right)^2}\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{3}{2}\\2{x^2} - 4x + 9 = 4{x^2} - 12x + 9\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{3}{2}\\2{x^2} - 8x = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{3}{2}\\2x\left( {x - 4} \right) = 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{3}{2}\\x = 0,x = 4\end{array} \right. \Leftrightarrow x = 4\)
Vậy phương trình có nghiệm \(x = 4\).
Chọn D.