Đáp án:
$\left[ \begin{array}{l}
f\left( x \right) > 0\,\,\,khi\,\,\,x \in \left( { - 1;\,\,\frac{3}{2}} \right) \cup \left( {2; + \infty } \right)\\
f\left( x \right) = 0\,\,\,khi\,\,\,\,x \in \left\{ { - 1;\,\,\frac{3}{2};\,\,2} \right\}\\
f\left( x \right) < 0\,\,\,\,khi\,\,\,x \in \left( { - \infty ; - 1} \right) \cup \left( {\frac{3}{2};\,\,2} \right)
\end{array} \right..$
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}f\left( x \right) = \left( {2{x^2} - x - 3} \right)\left( {x - 2} \right) = \left( {2x - 3} \right)\left( {x + 1} \right)\left( {x - 2} \right).\\ \Rightarrow f\left( x \right) = 0 \Leftrightarrow \left( {2x - 3} \right)\left( {x + 1} \right)\left( {x - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}2x - 3 = 0\\x + 1 = 0\\x - 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{3}{2}\\x = - 1\\x = 2\end{array} \right..\end{array}\)
Ta có bảng xét dấu như hình vẽ.
Vậy \(\left[ \begin{array}{l}f\left( x \right) > 0\,\,\,khi\,\,\,x \in \left( { - 1;\,\,\frac{3}{2}} \right) \cup \left( {2; + \infty } \right)\\f\left( x \right) = 0\,\,\,khi\,\,\,\,x \in \left\{ { - 1;\,\,\frac{3}{2};\,\,2} \right\}\\f\left( x \right) < 0\,\,\,\,khi\,\,\,x \in \left( { - \infty ; - 1} \right) \cup \left( {\frac{3}{2};\,\,2} \right)\end{array} \right..\)
