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Ta có:
$\begin{array}{l}
(1) \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 1\\
5{x^2} - 8x + 3 > {x^2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(a)\\
\left\{ \begin{array}{l}
0 < x < 1\\
0 < 5{x^2} - 8x + 3 < {x^2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,(q)
\end{array} \right.\\
(a) \Leftrightarrow \left\{ \begin{array}{l}
x > 1\\
4{x^2} - 8x + 3 > 0
\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x > 1\\
x < \frac{1}{2}\,\,\,\,;\,\,\,x > \frac{3}{2}
\end{array} \right.\,\,\,\,\,\, \Leftrightarrow x > \frac{3}{2}\\
(q) \Leftrightarrow \left\{ \begin{array}{l}
0 < x < 1\\
5{x^2} - 8x + 3 > 0\\
4{x^2} - 8x + 3 < 0
\end{array} \right.\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
0 < x < 1\\
x < \frac{3}{5};\,\,\,\,x > 1\,\,\,\, \Leftrightarrow \frac{1}{2} < x < \frac{3}{5}\\
\frac{1}{2} < x < \frac{3}{2}
\end{array} \right.
\end{array}$
Nghiệm của bất phương trình ($1$) là :$\left[ \begin{array}{l}
\frac{1}{2} < x < \frac{3}{5}\\
x > \frac{3}{2}
\end{array} \right.$
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