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Chia $2$ vế của $(1)$ cho ${4^{\frac{1}{x}}}$ta có :
$\begin{array}{l}
x \ne 0\,\,\,\,\& \,\,(1) \Leftrightarrow 6{\left( {\frac{9}{4}} \right)^{\frac{1}{x}}} - 13{\left( {\frac{6}{4}} \right)^{\frac{1}{x}}} + 6 \le 0\\
\Leftrightarrow 6{\left( {\frac{3}{2}} \right)^{2\left( {\frac{1}{x}} \right)}} - 13{\left( {\frac{3}{2}} \right)^{\frac{1}{x}}} + 6 \le 0
\end{array}$
Đặt $t = {\left( {\frac{3}{2}} \right)^{\frac{1}{x}}},\,\,t > 0$ ta có :
$\begin{array}{l}
6{t^2} - 13t + 6 \le 0 \Leftrightarrow \frac{2}{3} \le t \le \frac{3}{2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {\left( {\frac{3}{
2}} \right)^{ - 1}} \le {\left( {\frac{3}{2}} \right)^{\frac{1}{x}}} \le \frac{3}{2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow - 1 \le \frac{1}{x} \le 1\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
x \le - 1\\
x \ge 1
\end{array} \right.
\end{array}$
Vậy $x\geq 1$ hoặc $x\leq -1$ thỏa mãn đề bài.
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Đăng bài 26-04-12 08:47 AM
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