Đáp án:
\(
\left[ {\begin{array}{*{20}c}
{x = - 88} \\
{x = - 24} \\
{x = 3} \\
\end{array}} \right.
\)
Giải thích các bước giải:
\(
\begin{array}{l}
\sqrt[3]{{24 + x}} + \sqrt {12 - x} = 6 \\
Đk:x \le 12 \\
Đặt:\left\{ {\begin{array}{*{20}c}
{\sqrt[3]{{24 + x}} = u} \\
{\sqrt {12 - x} = v} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{24 + x = u^3 } \\
{12 - x = v^2 } \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{x = u^3 - 24} \\
{x = 12 - v^2 } \\
\end{array}} \right. \\
= > \left\{ {\begin{array}{*{20}c}
{u + v = 6} \\
{u^3 - 24 = 12 - v^2 } \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{v = 6 - u} \\
{u^3 + (6 - u)^2 = 36} \\
\end{array}} \right. \\
\Leftrightarrow \left\{ {\begin{array}{*{20}c}
{v = 6 - u} \\
{u^3 + u^2 - 12u = 0} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{v = 6 - u} \\
{\left[ {\begin{array}{*{20}c}
{u = - 4} \\
{u = 0} \\
{u = 3} \\
\end{array}} \right.} \\
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}c}
{\left\{ {\begin{array}{*{20}c}
{u = - 4} \\
{v = 10} \\
\end{array}} \right.} \\
{\left\{ {\begin{array}{*{20}c}
{u = 0} \\
{v = 6} \\
\end{array}} \right.} \\
{\left\{ {\begin{array}{*{20}c}
{u = 3} \\
{v = 3} \\
\end{array}} \right.} \\
\end{array}} \right. \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{x = - 88} \\
{x = - 24} \\
{x = 3} \\
\end{array}} \right.(tm) \\
\end{array}
\)
