ข้อสอบคณิต 2 - มีนาคม 2563

$$\begin{gathered} \mathrm{จากโจทย์} \\ {\left(-\frac{1}{32}\right)}^{-\frac{3}{5}} \\ \mathrm{จะได้} \frac{1}{{\left(-{\displaystyle \frac{1}{32}}\right)}^{{\displaystyle \frac{3}{5}}}} = \frac{1}{\sqrt[5]{{\left(-{\displaystyle \frac{1}{32}}\right)}^{{\displaystyle 3}}}} \end{gathered}$$

$$\begin{gathered} = \frac{1}{\sqrt[5]{{\left[{\left(-{\displaystyle \frac{1}{2}}\right)}^5\right]}^3}} \\ = \frac{1}{{\left(-{\displaystyle \frac{1}{2}}\right)}^3} \end{gathered}$$

$= \frac{1}{-{\displaystyle \frac{1}{8}}} = -8 \to \mathrm{ตอบ} 1$