Đáp án đúng: C
Giải chi tiết:\(\left\{ \begin{array}{l}3\sqrt {x - 1} - \sqrt {y + 2} = 5\\2\sqrt {x - 1} + 3\sqrt {y + 2} = 18\end{array} \right.\)
Điều kiện: \(x \ge 1;\,\,y \ge - 2.\)
Đặt \(\left\{ \begin{array}{l}\sqrt {x - 1} = a\,\,\,\,\,\,\left( {a \ge 0} \right)\\\sqrt {y + 2} = b\,\,\,\,\,\left( {b \ge 0} \right)\end{array} \right..\,\,\,\,\) Khi đó ta có hệ phương trình:
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}3a - b = 5\\2a + 3b = 18\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}9a - 3b = 15\\2a + 3b = 18\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}11a = 33\\b = 3a - 5\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a = 3\,\,\,\,\left( {tm} \right)\\b = 3a - 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = 4\,\,\,\left( {tm} \right)\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}\sqrt {x - 1} = 3\\\sqrt {y + 2} = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x - 1 = 9\\y + 2 = 16\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 10\,\,\,\left( {tm} \right)\\y = 14\,\,\,\left( {tm} \right)\end{array} \right.\\ \Rightarrow {y_0} - {x_0} = 4.\end{array}\)
Chọn C.
