Bài tập 4.31 trang 207 SBT Toán 12

a) \({x^3} - 8 = 0\)\( \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) = 0\) \( \Leftrightarrow \left[ \begin{array}{l}x - 2 = 0\\{x^2} + 2x + 4 = 0\end{array} \right.\)

\(\begin{array}{l}
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
{x^2} + 2x + 1 = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
{\left( {x + 1} \right)^2} = {\left( {i\sqrt 3 } \right)^2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 1 = i\sqrt 3 \\
x + 1 = - i\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - 1 + i\sqrt 3 \\
x = - 1 - i\sqrt 3
\end{array} \right.
\end{array}\)

b) \({x^3} + 8 = 0\)\( \Leftrightarrow \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) = 0\)\( \Leftrightarrow \left[ \begin{array}{l}x + 2 = 0\\{x^2} - 2x + 4 = 0\end{array} \right.\)

\(\begin{array}{l}
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
{x^2} - 2x + 1 = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
{\left( {x - 1} \right)^2} = {\left( {i\sqrt 3 } \right)^2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x - 1 = i\sqrt 3 \\
x - 1 = - i\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\\
x = 1 + i\sqrt 3 \\
x = 1 - i\sqrt 3
\end{array} \right.
\end{array}\)

-- Mod Toán 12 HỌC247