Đáp án:
B1:
\(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{1}{3}} \right) \cup \left( {1; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \dfrac{1}{3};1} \right)
\end{array}\)
Giải thích các bước giải:
B1:
BXD:
x -∞ -1/3 1 +∞
f(x) + 0 - 0 +
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - \dfrac{1}{3}} \right) \cup \left( {1; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - \dfrac{1}{3};1} \right)
\end{array}\)
\(\begin{array}{l}
B2:\\
Xét:g\left( x \right) = 0\\
\to 2{x^2} - 6x + 4 = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = 1
\end{array} \right.
\end{array}\)
BXD:
x -∞ 1 2 +∞
g(x) + 0 - 0 +
\(\begin{array}{l}
KL:g\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ;1} \right) \cup \left( {2; + \infty } \right)\\
g\left( x \right) < 0 \Leftrightarrow x \in \left( {1;2} \right)
\end{array}\)
\(\begin{array}{l}
B3:\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to h\left( x \right) \ge 0\forall x\\
B4:
\end{array}\)
BXD:
x -∞ 2(kép) 3(kép) 4 +∞
k(x) - 0 - 0 - 0 +
\(\begin{array}{l}
KL:k\left( x \right) > 0 \Leftrightarrow x \in \left( {4; + \infty } \right)\\
k\left( x \right) \le 0 \Leftrightarrow x \in \left( { - \infty ;4} \right]
\end{array}\)
