Đáp án:
1b) \(KL:x \in \left( { - \infty ; - \dfrac{3}{2}} \right) \cup \left( { - 1;1} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
C4:\\
\left( {x + 1} \right)\left( {x + 4} \right) < 5\sqrt {{x^2} + 5x + 28} \\
\to {x^2} + 5x + 4 < 5\sqrt {{x^2} + 5x + 28} \\
Đặt:\sqrt {{x^2} + 5x + 28} = t\left( {t \ge 0} \right)\\
\to {x^2} + 5x + 28 = {t^2}\\
\to {x^2} + 5x = {t^2} - 28\\
Bpt \to {t^2} - 28 + 4 < 5t\\
\to {t^2} - 5t - 24 < 0\\
\to \left( {t - 8} \right)\left( {t + 3} \right) < 0\\
\to - 3 < t < 8\\
\to 0 \le t < 8\\
\to \sqrt {{x^2} + 5x + 28} < 8\\
\to {x^2} + 5x + 28 < 64\\
\to {x^2} + 5x - 36 < 0\\
\to \left( {x - 4} \right)\left( {x + 9} \right) < 0\\
\to - 9 < x < 4\\
C1b)\\
\left\{ \begin{array}{l}
2{x^2} + 7x + 3 > 2x\\
\left( {1 - x} \right)\left( {x + 1} \right)\left( {3x + 4} \right) \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2{x^2} + 5x + 3 > 0\\
\left( {1 - x} \right)\left( {x + 1} \right)\left( {3x + 4} \right) \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {x + 1} \right)\left( {2x + 3} \right) > 0\left( 1 \right)\\
\left( {1 - x} \right)\left( {x + 1} \right)\left( {3x + 4} \right) \ge 0\left( 2 \right)
\end{array} \right.
\end{array}\)
BXD: (1)
x -∞ -1 -3/2 +∞
f(x) + 0 - 0 +
\( \to x \in \left( { - \infty ; - 1} \right) \cup \left( { - \dfrac{3}{2}; + \infty } \right)\)
BXD: (2)
x -∞ -4/3 -1 1 +∞
f(x) + 0 - 0 + 0 -
\( \to x \in \left( { - \infty ; - \dfrac{4}{3}} \right] \cup \left[ { - 1;1} \right]\)
\(KL:x \in \left( { - \infty ; - \dfrac{3}{2}} \right) \cup \left( { - 1;1} \right]\)
