Đáp án:
f) \(\begin{array}{l}
f\left( x \right) < 0 \to x \in \left( { - 7;\dfrac{3}{2}} \right) \cup \left( {5; + \infty } \right)\\
f\left( x \right) > 0 \to x \in \left( { - \infty ; - 7} \right) \cup \left( {\dfrac{3}{2};5} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
f)f\left( x \right) = 0\\
\to \left( {x - 5} \right)\left( {x + 7} \right)\left( {3 - 2x} \right) = 0\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 7\\
x = \dfrac{3}{2}
\end{array} \right.
\end{array}\)
BXD:
x -∞ -7 3/2 5 +∞
f(x) + 0 - 0 + 0 -
\(\begin{array}{l}
KL:f\left( x \right) < 0 \to x \in \left( { - 7;\dfrac{3}{2}} \right) \cup \left( {5; + \infty } \right)\\
f\left( x \right) > 0 \to x \in \left( { - \infty ; - 7} \right) \cup \left( {\dfrac{3}{2};5} \right)
\end{array}\)
\(\begin{array}{l}
g)f\left( x \right) = 0\\
\to \left( {6 - 2x} \right)\left( {4 - x} \right)\left( {4 + x} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = 4\\
x = - 4
\end{array} \right.
\end{array}\)
BXD:
x -∞ -4 3 4 +∞
f(x) - 0 + 0 - 0 +
\(\begin{array}{l}
h)f\left( x \right) = 0\\
\to \left( { - 3{x^2} + 16x - 5} \right)\left( {3{x^2} - 3x + 7} \right) = 0\\
\to \left( {3x + 1} \right)\left( {5 - x} \right) = 0\left( {do:3{x^2} - 3x + 7 > 0\forall x} \right)\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{3}\\
x = 5
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1/3 5 +∞
f(x) - 0 + 0 -
\(\begin{array}{l}
i)f\left( x \right) = 0\\
\to \left( {x - 3} \right)\left( {3x + 1} \right)\left( {4x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{1}{3}\\
x = \dfrac{5}{4}
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1/3 5/4 3 +∞
f(x) - 0 + 0 - 0 +
\(\begin{array}{l}
k)f\left( x \right) = 0\\
\to x\left( {3x - 4} \right)\left( {2x + 1} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{4}{3}\\
x = - \dfrac{1}{2}\\
x = 1
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1/2 0 1 4/3 +∞
f(x) + 0 - 0 + 0 - 0 +
