Giải thích các bước giải:
Ta có;
\(\begin{array}{l}
a,\\
{\left( {x - 3} \right)^6} = x - 3\\
\Leftrightarrow {\left( {x - 3} \right)^6} - \left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right).\left[ {{{\left( {x - 3} \right)}^5} - 1} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
{\left( {x - 3} \right)^5} - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x - 3 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 4
\end{array} \right.\\
c,\\
{2020^{{x^2} - x}} = 1\\
\Leftrightarrow {x^2} - x = 0\\
\Leftrightarrow x\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
d,\\
{2^3} < {3^x} \le {27^3}\\
\Leftrightarrow 8 < {3^x} \le {\left( {{3^3}} \right)^3}\\
\Leftrightarrow 9 \le {3^x} \le {3^9}\\
\Leftrightarrow {3^2} \le {3^x} \le {3^9}\\
\Leftrightarrow 2 \le x \le 9\\
b,\\
{\left( {2x - 3} \right)^4} - 27\left( {2x - 3} \right) = 0\\
\Leftrightarrow \left( {2x - 3} \right).\left[ {{{\left( {2x - 3} \right)}^3} - 27} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 0\\
{\left( {2x - 3} \right)^3} - 27 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 0\\
{\left( {2x - 3} \right)^3} = {3^3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 0\\
2x - 3 = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 3\\
2x = 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = 3
\end{array} \right.
\end{array}\)
