Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
c.ĐK:x \ne 1;x \ne \frac{3}{2}\\
2{x^2} - 3x + 2x - 3 + 6{x^2} - 6x = 16{x^2} - 24x - 16x + 24\\
\to x = \frac{{13 \pm \sqrt {61} }}{6}\left( {TM} \right)\\
d.x - 4 = \sqrt {2x + 16} \\
\to \left\{ \begin{array}{l}
x \ge 4\\
{x^2} - 8x + 16 = 2x + 16
\end{array} \right. \to \left\{ \begin{array}{l}
x \ge 4\\
\left[ \begin{array}{l}
x = 10\left( {TM} \right)\\
x = 0\left( l \right)
\end{array} \right.
\end{array} \right.\\
e.3x - 4 = \sqrt { - {x^2} + 3x + 2} \\
\to \left\{ \begin{array}{l}
x \ge \frac{4}{3}\\
9{x^2} - 24x + 16 = - {x^2} + 3x + 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge \frac{4}{3}\\
\left[ \begin{array}{l}
x = 2\left( {TM} \right)\\
x = \frac{7}{{10}}\left( l \right)
\end{array} \right.
\end{array} \right.\\
b.ĐK:x \ne 2\\
Pt \to 4x - 8 + {x^2} - 2x = 3x - 2\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.\left( {TM} \right)
\end{array}\)
