Đáp án:
b) \(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 3} \right) \cup \left( {1;2} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - 3;1} \right) \cup \left( {2; + \infty } \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C2:\\
a)4 - 25{x^2} = \left( {2 - 5x} \right)\left( {2 + 5x} \right)
\end{array}\)
BXD:
x -∞ -2/5 2/5 +∞
f(x) - 0 + 0 -
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \dfrac{2}{5};\dfrac{2}{5}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{2}{5}} \right) \cup \left( {\dfrac{2}{5}; + \infty } \right)
\end{array}\)
\(\begin{array}{l}
b)f\left( x \right) = - {x^3} + 7x - 6\\
= \left( {x + 3} \right)\left( {2 - x} \right)\left( {x - 1} \right)
\end{array}\)
BXD:
x -∞ -3 1 2 +∞
f(x) + 0 - 0 + 0 -
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 3} \right) \cup \left( {1;2} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - 3;1} \right) \cup \left( {2; + \infty } \right)
\end{array}\)
\(\begin{array}{l}
c)f\left( x \right) = {x^2} - x - 2\sqrt 2 \\
\Delta = 1 + 4\sqrt 2 \\
\to \left\{ \begin{array}{l}
{x_1} = \dfrac{{1 + \sqrt {1 + 4\sqrt 2 } }}{2}\\
{x_2} = \dfrac{{1 - \sqrt {1 + 4\sqrt 2 } }}{2}
\end{array} \right.
\end{array}\)
BXD:
x -∞ \({x_2}\) \({x_1}\) +∞
f(x) + 0 - 0 +
