Đáp án:
a) \(x \in \left[ { - 7;1} \right] \cup \left[ {2; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Xét:\left( {4 - 2x} \right)\left( {{x^2} + 6x - 7} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = 1\\
x = - 7
\end{array} \right.
\end{array}\)
BXD:
x -∞ -7 1 2 +∞
f(x) + 0 - 0 + 0 -
\(KL:x \in \left[ { - 7;1} \right] \cup \left[ {2; + \infty } \right)\)
\(\begin{array}{l}
b)DK:x \ne \left\{ { - 3; - \dfrac{7}{3};1;3} \right\}\\
\dfrac{{3{x^2} + 4x - 7 - 3\left( {{x^2} - 9} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)\left( {3x + 7} \right)\left( {x - 1} \right)}} < 0\\
\to \dfrac{{4x + 20}}{{\left( {x + 3} \right)\left( {x - 3} \right)\left( {3x + 7} \right)\left( {x - 1} \right)}} < 0
\end{array}\)
BXD:
x -∞ -5 -3 -7/3 1 3 +∞
f(x) - 0 + // - // + // - // +
\(KL:x \in \left( { - \infty ; - 5} \right) \cup \left( { - 3; - \dfrac{7}{3}} \right) \cup \left( {1;3} \right)\)
