Đáp án đúng: A
Giải chi tiết:ĐK: \(\left\{ \begin{array}{l}x + 1 \ge 0\\4 - x \ge 0\\ - {x^2} + 3x + 4 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge - 1\\x \le 4\\- 1 \le x \le 4\end{array} \right. \Leftrightarrow - 1 \le x \le 4.\)
\(\begin{array}{l}\,\,\,\,\,\,\,\sqrt {x + 1} + \sqrt {4 - x} + \sqrt { - {x^2} + 3x + 4} = 5\,\,\,\,\,\,\,\left( * \right)\\ \Leftrightarrow \sqrt {x + 1} + \sqrt {4 - x} + \sqrt {\left( {x + 1} \right)\left( {4 - x} \right)} = 5\end{array}\)
Đặt \(t = \sqrt {x + 1} + \sqrt {4 - x} \left( {t \ge 0} \right)\)
\(\begin{array}{l} \Rightarrow {t^2} = x + 1 + 4 - x + 2\sqrt {\left( {x + 1} \right)\left( {4 - x} \right)} \Leftrightarrow \sqrt {\left( {x + 1} \right)\left( {4 - x} \right)} = \frac{{{t^2} - 5}}{2}\\\Rightarrow \left( * \right) \Leftrightarrow t + \frac{{{t^2} - 5}}{2} = 5\\\Leftrightarrow {t^2} + 2t - 15 = 0\\ \Leftrightarrow {t^2} + 2t + 1 - 16 = 0\\ \Leftrightarrow {\left( {t + 1} \right)^2} = 16 \Leftrightarrow \left[ \begin{array}{l}t + 1 = 4\\t + 1 = - 4\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}t = 3\,\,\,\,\left( {tm} \right)\\t = - 5\,\,\,\left( {ktm} \right)\end{array} \right..\end{array}\)
Với \(t = 3\) ta có:
\(\begin{array}{l}\,\,\,\,\,\,\,\sqrt {x + 1} + \sqrt {4 - x} = 3\\ \Leftrightarrow x + 1 + 2\sqrt {\left( {x + 1} \right)\left( {4 - x} \right)} + 4 - x = 9\\\Leftrightarrow \sqrt { - {x^2} + 3x + 4} = 2\\ \Leftrightarrow - {x^2} + 3x + 4 = 4\\\Leftrightarrow - {x^2} + 3x = 0\\ \Leftrightarrow x\left( {3 - x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\3 - x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\,\,\,\,\,\left( {tm} \right)\\x = 3\,\,\,\,\,\left( {tm} \right)\end{array} \right..\end{array}\)
Vậy phương trình có hai nghiệm: \(x = 0\) và \(x = 3.\)
Chọn A .
