Tìm \(x\) biết \({x^3}\;-12{x^2}\; + 48x-64 = 0\)
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A.
\(x =- 4\).
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B.
\(x = 4\).
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C.
\(x =- 8\).
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D.
\(x = 8\).
\(\begin{array}{l}{x^3}\;-12{x^2}\; + 48x-64 = 0 \Leftrightarrow {x^3}\;-{{ 3}}.{x^2}.4 + 3.x{.4^2} - {4^3} = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {\left( {x - 4} \right)^3} = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow x - 4 = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow x = 4\end{array}\)
Đáp án : B