Đáp án đúng: A
Giải chi tiết:\(\sqrt{x+4}+\sqrt{x+1}=\sqrt{2x+9}\)
Điều kiện:\(\left\{ \begin{array}{l}x + 4 \ge 0\\x + 1 \ge 0\\2x + 9 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge - 4\\x \ge - 1\\x \ge \frac{{ - 9}}{2}\end{array} \right. \Leftrightarrow x \ge - 1\)
\(\begin{array}{l}PT \Leftrightarrow (x + 4) + 2\sqrt {(x + 4)(x + 1)} + (x + 1) = 2x + 9\\\,\,\,\,\,\,\, \Leftrightarrow 2\sqrt {(x + 4)(x + 1)} = 4\\\,\,\,\,\,\,\, \Leftrightarrow \sqrt {(x + 4)(x + 1)} = 2\\\,\,\,\,\,\,\, \Leftrightarrow (x + 4)(x + 1) = 4\\\,\,\,\,\,\,\, \Leftrightarrow {x^2} + 5x + 4 = 4\\\,\,\,\,\,\,\, \Leftrightarrow {x^2} + 5x = 0\\\,\,\,\,\,\,\, \Leftrightarrow x(x + 5) = 0\\\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}x = 0\\x + 5 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\,\,\,\,\,\,(tm)\\x = - 5\,\,\,(ktm)\end{array} \right.\end{array}\)
Vậy phương trình có nghiệm duy nhất x = 0.
Chọn A.