Đáp án:
\(x \in \left( { - \infty ;0} \right) \cup \left[ {1;\dfrac{3}{2}} \right]\)
Giải thích các bước giải:
\(\begin{array}{l}
6)DK:x \le 2\\
\dfrac{{\sqrt {2 - x} + 4x - 3 - 2x}}{x} \ge 0\\
\to \dfrac{{\sqrt {2 - x} + 2x - 3}}{x} \ge 0\\
Xét:\sqrt {2 - x} + 2x - 3 = 0\\
\to \sqrt {2 - x} = 3 - 2x\\
\to 2 - x = 9 - 12x + 4{x^2}\left( {DK:\dfrac{3}{2} \ge x} \right)\\
\to 4{x^2} - 11x + 7 = 0\\
\to \left( {4x - 7} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{4}\left( l \right)\\
x = 1
\end{array} \right.
\end{array}\)
BXD:
x -∞ 0 1 3/2
f(x) + // - 0 +
\(KL:x \in \left( { - \infty ;0} \right) \cup \left[ {1;\dfrac{3}{2}} \right]\)
