Toán Lớp 12: Giải phương trình : $(log_{3}3x)^2$ $-$ $4log_{3}x$ $-$ $4$ $=$ $0$
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Toán Lớp 12: Giải phương trình : $(log_{3}3x)^2$ $-$ $4log_{3}x$ $-$ $4$ $=$ $0$
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⇔ (log_{3} + t)^2 – 4t = 4 = 0
\quad \left(\log_33x\right)^2 – 4\log_3x – 4 =0\qquad (ĐK:x >0)\\
\Leftrightarrow \left(1 + \log_3x\right)^2 – 4\log_3x – 4 =0\\
\Leftrightarrow \log_3^2x + 2\log_3x + 1 – 4\log_3x – 4 =0\\
\Leftrightarrow \log_3^2x – 2\log_3x – 3 =0\\
\Leftrightarrow \left[\begin{array}{l}\log_3x = -1\\\log_3x = 3\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x =\dfrac13\\x = 27\end{array}\right.\quad (nhận)\\
\text{Vậy}\ S = \left\{\dfrac13;27\right\}
\end{array}\)