Đáp án:
d) \(x \in \left( { - \infty ; - 3} \right) \cup \left( { - \dfrac{1}{2};\dfrac{1}{2}} \right) \cup \left( {4; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Xét:3{x^2} - 8x + 2 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{4 + \sqrt {10} }}{3}\\
x = \dfrac{{4 - \sqrt {10} }}{3}
\end{array} \right.
\end{array}\)
BXD:
x -∞ \(\dfrac{{4 - \sqrt {10} }}{3}\) \(\dfrac{{4 + \sqrt {10} }}{3}\) +∞
f(x) + 0 - 0 +
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ;\dfrac{{4 - \sqrt {10} }}{3}} \right) \cup \left( {\dfrac{{4 + \sqrt {10} }}{3}; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( {\dfrac{{4 - \sqrt {10} }}{3};\dfrac{{4 + \sqrt {10} }}{3}} \right)
\end{array}\)
\(\begin{array}{l}
c)DK:x \ne \pm 3\\
Xét:2{x^2} - 3x + 1 = 0\\
\to \left( {2x - 1} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{2}
\end{array} \right.
\end{array}\)
BXD:
x -∞ -3 1/2 1 3 +∞
f(x) + // - 0 + 0 + // -
\(\begin{array}{l}
b)DK:x \ne \left\{ { - 3;2;3} \right\}\\
Xét:\left( {{x^2} - 4x + 3} \right){\left( {3 - 2x} \right)^2} = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 3\\
x = \dfrac{3}{2}
\end{array} \right.
\end{array}\)
BXD:
x -∞ -3 1 3/2(kép) 2 3 +∞
f(x) + // - 0 + 0 + // - // +
\(KL:x \in \left( { - 3;1} \right] \cup \left( {2;3} \right)\)
d) Xét:
\(\begin{array}{l}
\left( {4{x^2} - 1} \right)\left( { - {x^2} + x + 12} \right) > 0\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{1}{2}\\
x = 4\\
x = - 3
\end{array} \right.
\end{array}\)
BXD:
x -∞ -3 -1/2 1/2 4 +∞
f(x) - 0 + 0 - 0 + 0 -
\(KL:x \in \left( { - \infty ; - 3} \right) \cup \left( { - \dfrac{1}{2};\dfrac{1}{2}} \right) \cup \left( {4; + \infty } \right)\)
