Đáp án:
a) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 3\\
\sqrt {x - 3} = 13 - 3x\\
\to x - 3 = 169 - 78x + 9{x^2}\left( {DK:\dfrac{{13}}{3} \ge x} \right)\\
\to 9{x^2} - 79x + 172 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{43}}{9}\left( l \right)\\
x = 4\left( {TM} \right)
\end{array} \right.\\
b) - {x^2} + 3x - 3\sqrt {{x^2} - 3x + 4} = - 12\\
\to - 3\sqrt {{x^2} - 3x + 4} = {x^3} - 3x - 12\\
Đặt:\sqrt {{x^2} - 3x + 4} = t\left( {t \ge 0} \right)\\
\to {x^2} - 3x + 4 = {t^2}\\
\to {x^2} - 3x = {t^2} - 4\\
Pt \to - 3t = {t^2} - 4 - 12\\
\to {t^2} + 3t - 16 = 0\\
\to \left[ \begin{array}{l}
t = \dfrac{{ - 3 + \sqrt {73} }}{2}\\
t = \dfrac{{ - 3 - \sqrt {73} }}{2}\left( l \right)
\end{array} \right.\\
\to {x^2} - 3x + 4 = \dfrac{{41 - 3\sqrt {73} }}{2}\\
\to \left[ \begin{array}{l}
x = 3,935979142\\
x = - 0,9359791424
\end{array} \right.
\end{array}\)
