a) \(f(x) =(3{x^2} - 10x + 3)(4x - 5)\)
Ta có: \(f\left( x \right) = 0 \)
\(\Leftrightarrow \left( {3{x^2} - 10x + 3} \right)\left( {4x - 5} \right) = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}3{x^2} - 10x + 3 = 0\\4x - 5 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left( {3x - 1} \right)\left( {x - 3} \right) = 0\\4x - 5 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{1}{3}\\x = 3\\x = \frac{5}{4}\end{array} \right..\)
Ta có bảng xét dấu:
Kết luận:
\(f(x) < 0\) với \(x \in \left( { - \infty ;{1 \over 3}} \right) \cup \left( {{5 \over 4};3} \right)\)
\(f(x) > 0\) với \(x \in \left( {{1 \over 3};{5 \over 4}} \right) \cup \left( {3; + \infty } \right)\)
b) \(f(x) = (3{x^2} - 4x)(2{x^2} - x - 1)=0\)
\( \Leftrightarrow \left[ \begin{array}{l}3{x^2} - 4x = 0\\2{x^2} - x - 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x\left( {3x - 4} \right) = 0\\\left( {2x + 1} \right)\left( {x - 1} \right) = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = \frac{4}{3}\\x = - \frac{1}{2}\\x = 1\end{array} \right..\)
Ta có bảng xét dấu:
Vậy \(f\left( x \right) > 0\;\;khi\;\;x \in \left( { - \infty ;\; - \frac{1}{2}} \right)\)\( \cup \left( {0;\;1} \right) \cup \left( {\frac{4}{3}; + \infty } \right).\)
\(f\left( x \right) < 0\;\;khi\;\;x \in \left( { - \frac{1}{2};\;0} \right) \cup \left( {1;\;\frac{4}{3}} \right).\)
c) \(f(x) =\)\( (4{x^2} - 1)( - 8{x^2} + x - 3)(2x + 9)=0\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}4{x^2} - 1 = 0\\ - 8{x^2} + x - 3 = 0\\2x + 9 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left( {2x - 1} \right)\left( {2x + 1} \right) = 0\\ - 8\left( {{x^2} - \frac{1}{8}x} \right) - 3 = 0\\2x + 9 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x - 1 = 0\\2x + 1 = 0\\ - 8\left[ {{{\left( {x - \frac{1}{{16}}} \right)}^2} - \frac{1}{{256}}} \right] - 3 = 0\\2x + 9 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{1}{2}\\x = - \frac{1}{2}\\ - 8{\left( {x - \frac{1}{{16}}} \right)^2} - \frac{{95}}{{32}} = 0\;\;\left( {VN} \right)\\x = - \frac{9}{2}\end{array} \right..\end{array}\)
Ta có bảng xét dấu:
Vậy \(f\left( x \right) > 0\;\;khi\;\;x \in \left( { - \infty ; - \frac{9}{2}} \right) \)\(\cup \left( { - \frac{1}{2};\;\frac{1}{2}} \right).\)
\(f\left( x \right) < 0\;\;khi\;\;x \in \left( { - \frac{9}{2}; - \frac{1}{2}} \right)\)\( \cup \left( {\frac{1}{2}; + \infty } \right).\)
d) \(f(x) = \frac{(3x^{2}-x)(3-x^{2})}{4x^{2}+x-3}=0\)
\( \Leftrightarrow \left\{ \begin{array}{l}\left( {3{x^2} - x} \right)\left( {3 - {x^2}} \right) = 0\\4{x^2} + x - 3 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x\left( {3x - 1} \right)\left( {3 - {x^2}} \right) = 0\\\left( {3x - 4} \right)\left( {x + 1} \right) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 0\\x = \pm \sqrt 3 \\x = \frac{1}{3}\end{array} \right.\\x \ne - 1\\x \ne \frac{4}{3}\end{array} \right..\)
Ta có bảng xét dấu:
Vậy \(f\left( x \right) > 0\) khi \(x \in \left( { - \sqrt 3 ;\; - 1} \right) \cup \left( {0;\frac{1}{3}} \right) \cup \left( {\frac{3}{4};\;\sqrt 3 } \right).\)
\(f\left( x \right) < 0\) khi \(x \in \left( { - \infty ;\; - \sqrt 3 } \right) \cup \left( { - 1;0} \right) \cup \left( {\frac{1}{3};\;\frac{3}{4}} \right)\)\( \cup \left( {\sqrt 3 ; + \infty } \right).\)
