Đáp án:
\(\left[ \begin{array}{l}
x = 5\\
x = 0\\
x = - 3\\
x = - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK: - 5 \le x \le 5\\
x + \sqrt {25 - {x^2}} + x\sqrt {25 - {x^2}} = 5\\
\to \left( {x + 1} \right)\sqrt {25 - {x^2}} = 5 - x\\
\to \left( {x + 1} \right)\sqrt {\left( {5 - x} \right)\left( {5 + x} \right)} - \left( {5 - x} \right) = 0\\
\to \sqrt {5 - x} \left[ {\left( {x + 1} \right)\sqrt {5 + x} - \left( {\sqrt {5 - x} } \right)} \right] = 0\\
\to \left[ \begin{array}{l}
x = 5\\
\left( {{x^2} + 2x + 1} \right)\left( {5 + x} \right) = 5 - x
\end{array} \right.\\
\to {x^3} + 2{x^2} + x + 5{x^2} + 10x + 5 = 5 - x\\
\to {x^3} + 7{x^2} + 12x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
{x^2} + 7x + 12 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 3\\
x = - 4
\end{array} \right.\left( {TM} \right)\\
KL:\left[ \begin{array}{l}
x = 5\\
x = 0\\
x = - 3\\
x = - 4
\end{array} \right.
\end{array}\)
