Đáp án:
1) \(x \in \left( { - 5; - 2} \right) \cup \left( { - 1;1} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne \pm 1\\
\dfrac{{\left( {x - 2} \right)\left( {x + 1} \right) - \left( {x - 3} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} < \dfrac{{{x^2} + 4x + 15}}{{\left( {1 - x} \right)\left( {x + 1} \right)}}\\
\to \dfrac{{{x^2} - x - 2 - {x^2} + 4x - 3 + {x^2} + 4x + 15}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} < 0\\
\to \dfrac{{{x^2} + 7x + 10}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} < 0\\
\to \dfrac{{\left( {x + 2} \right)\left( {x + 5} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} < 0
\end{array}\)
BXD:
x -∞ -5 -2 -1 1 +∞
f(x) + 0 - 0 + // - // +
\(KL:x \in \left( { - 5; - 2} \right) \cup \left( { - 1;1} \right)\)
\(\begin{array}{l}
b)2{x^2} - x - 3 \le {x^2} + 3\\
\to {x^2} - x - 6 \le 0\\
\to \left( {x - 3} \right)\left( {x + 2} \right) \le 0\\
\to - 2 \le x \le 3\\
c)2{x^2} + 7x + 3 < 12\\
\to 2{x^2} + 7x - 9 < 0\\
\to \left( {2x + 9} \right)\left( {x - 1} \right) < 0\\
\to - \dfrac{9}{2} < x < 1
\end{array}\)
