Toán Lớp 9: (4x + 2)(x ² + 1) = 0
(x ² + 1)(x ² – 4x + 4) = 0
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Toán Lớp 9: (4x + 2)(x ² + 1) = 0
(x ² + 1)(x ² – 4x + 4) = 0
Comments ( 2 )
\left( {4x{\rm{ }} + {\rm{ }}2} \right)\left( {{{\rm{x}}^2}{\rm{ }} + {\rm{ }}1} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {4x{\rm{ }} + {\rm{ }}2} \right) = 0\\
\left( {{x^2}{\rm{ }} + {\rm{ }}1} \right) = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ – 1}}{2}\\
{x^2} = 1\,\left( {vo\,\,li} \right)
\end{array} \right.\\
\left( {x{\rm{ }}{\rm{ }} + {\rm{ }}1} \right)\left( {x{\rm{ }}{\rm{ }} – {\rm{ }}4x{\rm{ }} + {\rm{ }}4} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left( {x{\rm{ }}{\rm{ }} + {\rm{ }}1} \right){\left( {x – 2} \right)^2} = 0\\
\Leftrightarrow \left( {x{\rm{ }}{\rm{ }} + {\rm{ }}1} \right)\left( {x – 2} \right)\left( {x – 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x{\rm{ }}{\rm{ }} + {\rm{ }}1} \right) = 0\\
\left( {x – 2} \right) = 0
\end{array} \right.\, \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 1\left( {vo\,\,li} \right)\\
x = 2
\end{array} \right.
\end{array}\]