Đáp án:
b. \(x \in \left( { - \infty ; - 2} \right] \cup \left[ {1;5} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\sqrt {2 - x} - \sqrt {7 - x} + \sqrt { - 3 - 2x} > 0\\
\to \sqrt {2 - x} + \sqrt { - 3 - 2x} > \sqrt {7 - x} \\
\to 2 - x + - 3 - 2x + 2\sqrt {\left( {2 - x} \right)\left( { - 3 - 2x} \right)} > 7 - x\left( {DK:x \le - \dfrac{3}{2}} \right)\\
\to - 1 - 3x + 2\sqrt { - 6 - 4x + 3x + 2{x^2}} > 7 - x\\
\to 2\sqrt {2{x^2} - x - 6} > 8 + 2x\\
\to \sqrt {2{x^2} - x - 6} > 4 + x\\
\to 2{x^2} - x - 6 > 16 + 8x + {x^2}\left( {x \ge - 4} \right)\\
\to {x^2} - 9x - 22 > 0\\
\to \left( {x - 11} \right)\left( {x + 2} \right) > 0\\
\to x \in \left( { - 2;11} \right)\\
KL:x \in \left( { - 2; - \dfrac{3}{2}} \right]\\
b.\left( {{x^2} + 3x + 2} \right)\left( {5 - x} \right) \ge 0\\
\to \left( {x + 1} \right)\left( {x + 2} \right)\left( {5 - x} \right) \ge 0
\end{array}\)
BXD:
x -∞ -2 -1 5 +∞
f(x) + 0 - 0 + 0 -
\(KL:x \in \left( { - \infty ; - 2} \right] \cup \left[ {1;5} \right]\)
