Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {x + 1} \right).\left( {3 - x} \right) = 0\\
\to - {x^2} + 3x - x + 3 = 0\\
\to \left( {x - 3} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
b.2{x^2} - x - 4x + 2 = 0\\
\to \left( {x - 2} \right)\left( {2x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = \frac{1}{2}
\end{array} \right.\\
c. - 9{x^2} + 3x - 27x + 9 = 0\\
\to \left( {x + 3} \right)\left( {3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 3\\
x = \frac{1}{3}
\end{array} \right.\\
d. - {x^4} + 81{x^2} - {x^2} + 81 = 0\\
t = {x^2}\left( {t > 0} \right)\\
\to - {t^2} + 80t + 81 = 0\\
\to \left( {t - 81} \right)\left( {t + 1} \right) = 0\\
\to \left[ \begin{array}{l}
t = 81\\
t = - 1\left( l \right)
\end{array} \right. \to {x^2} = 81 \to x = \pm 9\\
e.x - 5 = \sqrt[5]{{32}} = 2\\
\to x = 7\\
h.\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
7 > x
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 2\\
x > 7
\end{array} \right.(l)
\end{array} \right. \to 2 < x < 7\\
i.\left[ \begin{array}{l}
x - 7 \le 3\\
x - 7 \ge - 3
\end{array} \right. \to \left[ \begin{array}{l}
x \le 10\\
x \ge 4
\end{array} \right.
\end{array}\)
