Do \(x < 0,y \ge 0\) nên
\(\begin{array}{l}3\sqrt {{x^2}y} + x\sqrt y \\ = 3\sqrt {{x^2}} .\sqrt y + x\sqrt y \\ = 3\left| x \right|.\sqrt y + x\sqrt y \end{array}\)
Mà \(x < 0\) nên \(\left| x \right| = - x\)
\(\begin{array}{l}3\left| x \right|.\sqrt y + x\sqrt y \\ = - 3x\sqrt y + x\sqrt y \\ = - 2x\sqrt y \end{array}\)
Vậy đáp án là (C).
