Giải phương trình:${{\log }_{2}}\left( {{x}^{2}}-x+\frac{9}{4} \right)={{\log }_{2}}x.{{\log }_{2}}4x$ (*)
Giải:
Điều kiện: $0<x\ne 2.$
$\ \begin{array}{l}
\left( * \right) \Leftrightarrow {\log _2}\left( {{x^2} - x + \frac{9}{4}} \right) + {\log _2}x.\left( {{{\log }_2}4 + {{\log }_2}x} \right) = 0\\
\Leftrightarrow {\log _2}\left( {{x^2} - x + \frac{9}{4}} \right) + {\log _2}x.\left( {2 + {{\log }_2}x} \right) = 0\\
\Leftrightarrow {\log _2}\left( {{x^2} - x + \frac{9}{4}} \right) + {\left( {{{\log }_2}x} \right)^2} + 2{\log _2}x = 0\\
\Leftrightarrow {\log _2}\left( {{x^2} - x + \frac{9}{4}} \right) - 1 + {\left( {{{\log }_2}x} \right)^2} + 2{\log _2}x + 1 = 0\\
\Leftrightarrow {\log _2}\frac{1}{2}\left( {{x^2} - x + \frac{9}{4}} \right) + {\left( {{{\log }_2}x + 1} \right)^2} = 0
\end{array}.$
Ta có: $\ \left\{ \begin{array}{l}
{x^2} - x + \frac{9}{4} = {\left( {x - \frac{1}{2}} \right)^2} + 2 \ge 2 \Leftrightarrow \frac{1}{2}\left( {{x^2} - x + \frac{9}{4}} \right) \ge 1 \Leftrightarrow {\log _2}\frac{1}{2}\left( {{x^2} - x + \frac{9}{4}} \right) \ge 0\\
{\left( {{{\log }_2}x + 1} \right)^2} \ge 0
\end{array} \right.$
$\ \Rightarrow {\log _2}\frac{1}{2}\left( {{x^2} - x + \frac{9}{4}} \right) + {\left( {{{\log }_2}x + 1} \right)^2} = 0 \Leftrightarrow \left\{ \begin{array}{l}
x - \frac{1}{2} = 0\\
{\log _2}x + 1 = 0
\end{array} \right. \Leftrightarrow x = \frac{1}{2} \Rightarrow S = \left\{ {\frac{1}{2}} \right\}.$
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