Đáp án:
a. x=0
b. x=-1/4
c. x=-4
d. x=4
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \left\{ {5;8} \right\}\\
Pt \to 6x - 48 + 2x - 10 + 18 + {x^2} - 13x + 40 = 0\\
\to {x^2} - 5x = 0\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
x = 5\left( l \right)
\end{array} \right.\\
b.DK:x \ne 1\\
Pt \to {x^2} + x + 1 - 3{x^2} - 2{x^2} + 2x = 0\\
\to - 4{x^2} + 3x + 1 = 0\\
\to \left( {4x + 1} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = - \frac{1}{4}\left( {TM} \right)\\
x = 1\left( l \right)
\end{array} \right.\\
c.DK:x \ne \pm 3;x \ne - \frac{7}{2}\\
Pt \to 13x + 39 + {x^2} - 9 - 12x - 42 = 0\\
\to {x^2} + x - 12 = 0\\
\to \left( {x - 3} \right)\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\left( l \right)\\
x = - 4\left( {TM} \right)
\end{array} \right.\\
d.DK:x \ne \pm 2\\
Pt \to 3{x^2} + 2x - 8 - 4x - 2{x^2} + 4x = 0\\
\to {x^2} + 2x - 8 = 0\\
\to \left( {x - 2} \right)\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = 4\left( {TM} \right)
\end{array} \right.
\end{array}\)
