ข้อสอบคณิต O-NET - 2549

$$\begin{gathered} \mathrm{จากโจทย์} \frac{\sqrt[5]{-32}}{\sqrt[3]{27}}+\frac{2^6}{{\left(64\right)}^{{\displaystyle \frac{3}{2}}}} \\ \mathrm{จะได้} \frac{\sqrt[5]{\left(-2\right)·\left(-2\right)·\left(-2\right)·\left(-2\right)·\left(-2\right)}}{\sqrt[3]{3·3·3}}+\frac{2^6}{{\left(8·8\right)}^{{\displaystyle \frac{3}{2}}}} \\ = \frac{{\left[{\left(-2\right)}^{\cancel{5}}\right]}^{{\displaystyle \frac{1}{\cancel{5}}}}}{{\left[{\left(3\right)}^{\cancel{3}}\right]}^{{\displaystyle \frac{1}{\cancel{3}}}}}+\frac{2^6}{{\left[{\left(8\right)}^{\cancel{2}}\right]}^{{\displaystyle \frac{3}{\cancel{2}}}}} \end{gathered}$$

$$\begin{gathered} = \frac{\left(-2\right)}{3}+\frac{2^6}{8^3} \\ = \frac{-2}{3}+\frac{2^6}{{\left(2^3\right)}^3} = \frac{-2}{3}+\frac{2^6}{2^9} \\ = \frac{-2}{3}+\frac{1}{2^3} \\ = \frac{-16+3}{24} \\ = \frac{-13}{24} \\ \mathrm{ดังนั้น} \mathrm{ตอบข้อ} 1 \end{gathered}$$