Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
a){\rm{ }}2.\left( {x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} - {\rm{ }}x.\left( {x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
vay:S = \{ - 3;2\} \\
b){x^3} + 3{x^3} - 3x - 1 = 0\\
\Leftrightarrow 4{x^3} - 3x - 1 = 0\\
\Leftrightarrow 4{x^3} - 4{x^2} + 4{x^2} - 4x + x - 1 = 0\\
\Leftrightarrow 4{x^2}\left( {x - 1} \right) + 4x\left( {x - 1} \right) + \left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {4{x^2} + 4x + 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right){\left( {2x + 1} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
2x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{ - 1}}{2}
\end{array} \right.\\
vay:S = \{ 1;\frac{{ - 1}}{2}\} \\
c)2{x^2} - 18 = 0\\
\Leftrightarrow 2x\left( {x - 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 0\\
x - 9 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 9
\end{array} \right.\\
vay:S = \{ 0;9\}
\end{array}\]

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Answer 2

\[\begin{array}{l}
a){\rm{ }}2.\left( {x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} - {\rm{ }}x.\left( {x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} = {\rm{ }}0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
vay:S = \{ - 3;2\} \\
b){x^3} + 3{x^3} - 3x - 1 = 0\\
\Leftrightarrow 4{x^3} - 3x - 1 = 0\\
\Leftrightarrow 4{x^3} - 4{x^2} + 4{x^2} - 4x + x - 1 = 0\\
\Leftrightarrow 4{x^2}\left( {x - 1} \right) + 4x\left( {x - 1} \right) + \left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {4{x^2} + 4x + 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right){\left( {2x + 1} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
2x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \frac{{ - 1}}{2}
\end{array} \right.\\
vay:S = \{ 1;\frac{{ - 1}}{2}\} \\
c)2{x^2} - 18 = 0\\
\Leftrightarrow 2x\left( {x - 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 0\\
x - 9 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 9
\end{array} \right.\\
vay:S = \{ 0;9\}
\end{array}\]

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