Đáp án:
4d) \(x \in \left( { - \infty ; - 8} \right) \cup \left( { - \dfrac{5}{2}; - \dfrac{5}{3}} \right) \cup \left( {\dfrac{3}{2}; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
3f)DK:x \ne \left\{ { - 4;1} \right\}\\
f\left( x \right) = \dfrac{{{x^2} - 4{x^2} + 24x - 36}}{{\left( {1 - x} \right)\left( {x + 4} \right)}}\\
= \dfrac{{ - 3{x^2} + 24x - 36}}{{\left( {1 - x} \right)\left( {x + 4} \right)}}\\
= \dfrac{{-3\left( {x+6} \right)\left( {x - 2} \right)}}{{\left( {1 - x} \right)\left( {x + 4} \right)}}
\end{array}\)
BXD:
x -∞ -6 -4 1 2 +∞
f(x) + // - 0 + // - 0 +
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 6} \right) \cup \left( { - 4;1} \right) \cup \left( {2; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { - 6; - 4} \right) \cup \left( {1;2} \right)
\end{array}\)
\(\begin{array}{l}
4d)\dfrac{{{{\left( {2x - 3} \right)}^2} - {{\left( {3x + 5} \right)}^2}}}{{\left( {3x + 5} \right)\left( {2x - 3} \right)}} < 0\\
\to \dfrac{{4{x^2} - 12x + 9 - 9{x^2} - 30x - 25}}{{\left( {3x + 5} \right)\left( {2x - 3} \right)}} < 0\\
\to \dfrac{{ - 5{x^2} - 42x - 16}}{{\left( {3x + 5} \right)\left( {2x - 3} \right)}} < 0\\
\to \dfrac{{\left( {2 + 5x} \right)\left( { - 8 - x} \right)}}{{\left( {3x + 5} \right)\left( {2x - 3} \right)}} < 0
\end{array}\)
BXD:
x -∞ -8 -5/3 -2/5 3/2 +∞
f(x) - 0 + 0 - // + // -
\(KL:x \in \left( { - \infty ; - 8} \right) \cup \left( { - \dfrac{5}{3}; - \dfrac{2}{5}} \right) \cup \left( {\dfrac{3}{2}; + \infty } \right)\)
