Đáp án đúng: B
Giải chi tiết:Điều kiện xác định: \(x + 1 \ge 0\,\, \Rightarrow x \ge - 1\).
\(\begin{array}{l}\,\,\,\,\,\,\,{x^2} - 6\left( {x + 3} \right)\sqrt {x + 1} + 14x + 3\sqrt {x + 1} + 13 = 0\\ \Leftrightarrow \left( {{x^2} + 2x + 1} \right) - 6\left( {x + 1} \right)\sqrt {x + 1} + 12\left( {x + 1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 12\sqrt {x + 1} + 3\sqrt {x + 1} = 0\\ \Leftrightarrow {\left( {x + 1} \right)^2} - 6\left( {x + 1} \right)\sqrt {x + 1} + 12\left( {x + 1} \right) - 9\sqrt {x + 1} = 0\\ \Leftrightarrow \sqrt {x + 1} \left[ {{{\left( {\sqrt {x + 1} } \right)}^3} - 6{{\left( {\sqrt {x + 1} } \right)}^2} + 12\sqrt {x + 1} - 9} \right] = 0\\ \Leftrightarrow \left\{ \begin{array}{l}\sqrt {x + 1} = 0\\{\left( {\sqrt {x + 1} } \right)^3} - 6{\left( {\sqrt {x + 1} } \right)^2} + 12\sqrt {x + 1} - 9 = 0\,\,\,\,\,\,\left( 2 \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = - 1\,\,\left( {tmdk} \right)\\{\left( {\sqrt {x + 1} } \right)^3} - 6{\left( {\sqrt {x + 1} } \right)^2} + 12\sqrt {x + 1} - 9 = 0\,\,\,\,\,\,\left( 2 \right)\end{array} \right.\end{array}\)
Đặt \(\sqrt {x + 1} = a\,\,\,\left( {a \ge 0} \right)\) thì phương trình (2) trở thành:
\(\begin{array}{l}\,\,\,\,\,\,\,{a^3} - 6{a^2} + 12a - 9 = 0\\ \Leftrightarrow {a^3} - 3{a^2} - 3{a^2} + 9a + 3a - 9 = 0\\ \Leftrightarrow {a^2}\left( {a - 3} \right) - 3a\left( {a - 3} \right) + 3\left( {a - 3} \right) = 0\\ \Leftrightarrow \left( {a - 3} \right)\left( {{a^2} - 3a + 3} \right) = 0\\ \Leftrightarrow \left( {a - 3} \right)\left[ {{{\left( {a - \dfrac{3}{2}} \right)}^2} + \dfrac{3}{4}} \right] = 0\\ \Leftrightarrow a = 3\,\,\,\,\,\left( {do\,{{\left( {a - \dfrac{3}{2}} \right)}^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\,\,\,\,\forall a \ge 0} \right)\,\,\left( {tmdk\,a \ge 0} \right)\end{array}\)
\( \Rightarrow \sqrt {x + 1} = 3 \Leftrightarrow x + 1 = 9 \Leftrightarrow x = 8\,\,\,\left( {tmdk\,\,\,x \ge - 1} \right)\)
Vậy phương trình có hai nghiệm \(x = - 1\) và \(x = 8\).
