Question

Đã trả lời bởi chuyên gia

Answer 1

\(\begin{array}{l}
a,{x^3} = 5{x^2} - 4x\\
\Leftrightarrow {x^3} - 5{{\rm{x}}^2} + 4{\rm{x}} = 0\\
\Leftrightarrow x\left( {{x^2} - 5{\rm{x}} + 4} \right) = 0\\
\Leftrightarrow x\left( {{x^2} - x - 4{\rm{x}} + 4} \right) = 0\\
\Leftrightarrow x\left[ {x\left( {x - 1} \right) - 4\left( {x - 1} \right)} \right] = 0\\
\Leftrightarrow x\left( {x - 1} \right)\left( {x - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 1 = 0\\
x - 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1\\
x = 4
\end{array} \right.\\
Vay:S = \{ 0;1;4\} \\
b,{x^3} = 3x - 2\\
\Leftrightarrow {x^3} - 3{\rm{x}} + 2 = 0\\
\Leftrightarrow {x^3} - x - 2{\rm{x}} + 2 = 0\\
\Leftrightarrow x\left( {{x^2} - 1} \right) - 2\left( {x - 1} \right) = 0\\
\Leftrightarrow x\left( {x - 1} \right)\left( {x + 1} \right) - 2\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} - x + 2{\rm{x}} - 2} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left[ {x\left( {x - 1} \right) + 2\left( {x - 1} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 1} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2}\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.\\
vay:S = \{ 1; - 2\}
\end{array}\)

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Answer 2

\(\begin{array}{l}
a,{x^3} = 5{x^2} - 4x\\
\Leftrightarrow {x^3} - 5{{\rm{x}}^2} + 4{\rm{x}} = 0\\
\Leftrightarrow x\left( {{x^2} - 5{\rm{x}} + 4} \right) = 0\\
\Leftrightarrow x\left( {{x^2} - x - 4{\rm{x}} + 4} \right) = 0\\
\Leftrightarrow x\left[ {x\left( {x - 1} \right) - 4\left( {x - 1} \right)} \right] = 0\\
\Leftrightarrow x\left( {x - 1} \right)\left( {x - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 1 = 0\\
x - 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1\\
x = 4
\end{array} \right.\\
Vay:S = \{ 0;1;4\} \\
b,{x^3} = 3x - 2\\
\Leftrightarrow {x^3} - 3{\rm{x}} + 2 = 0\\
\Leftrightarrow {x^3} - x - 2{\rm{x}} + 2 = 0\\
\Leftrightarrow x\left( {{x^2} - 1} \right) - 2\left( {x - 1} \right) = 0\\
\Leftrightarrow x\left( {x - 1} \right)\left( {x + 1} \right) - 2\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} - x + 2{\rm{x}} - 2} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left[ {x\left( {x - 1} \right) + 2\left( {x - 1} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 1} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2}\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2
\end{array} \right.\\
vay:S = \{ 1; - 2\}
\end{array}\)

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