Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^3} - 3{x^2} + 3x = 1 + 2020{\left( {x - 1} \right)^2}\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 = 2020{\left( {x - 1} \right)^2}\\
\Leftrightarrow {\left( {x - 1} \right)^3} = 2020{\left( {x - 1} \right)^2}\\
\Leftrightarrow {\left( {x - 1} \right)^3} - 2020{\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2}.\left[ {\left( {x - 1} \right) - 2020} \right] = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 2021} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 2021 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2021
\end{array} \right.\\
c,\\
{x^3} - 5x = 0\\
\Leftrightarrow x\left( {{x^2} - 5} \right) = 0\\
\Leftrightarrow x\left( {x - \sqrt 5 } \right)\left( {x + \sqrt 5 } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - \sqrt 5 = 0\\
x + \sqrt 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt 5 \\
x = - \sqrt 5
\end{array} \right.
\end{array}\)
Em xem lại đề câu b, chỗ \(\left( {3{x^2} + 24} \right):\left( {x + 2} \right)\) nhé!
