ข้อสอบ PAT 1 - มีนาคม 2559

$$\begin{gathered} \mathrm{จากโจทย์} A = \frac{2}{\sqrt[4]{3}}-\sqrt[4]{3} \\ \mathrm{จากโจทย์} B = \frac{\sqrt{3}-{\displaystyle \frac{1}{\sqrt{3}}}}{\sqrt[4]{3}-\sqrt{{\displaystyle \frac{1}{\sqrt{3}}}}} = \frac{\sqrt{3}-{\displaystyle \frac{1}{\sqrt{3}}}}{\sqrt[4]{3}-\frac{1}{\sqrt[4]{3}}} \\ = \frac{3^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{1}{3^{{\displaystyle \frac{1}{2}}}}}}{3^{{\displaystyle \frac{1}{4}}}-{\displaystyle \frac{1}{3^{{\displaystyle \frac{1}{4}}}}}} = \frac{{\left(3^{{\displaystyle \frac{1}{4}}}\right)}^2-{\left({\displaystyle \frac{1}{3^{{\displaystyle \frac{1}{4}}}}}\right)}^2}{3^{{\displaystyle \frac{1}{4}}}-\frac{1}{3^{{\displaystyle \frac{1}{4}}}}} \end{gathered}$$

$$\begin{gathered} \mathrm{กำหนดให้} 3^{\frac{1}{4}} = x \\ \mathrm{จะได้} B = \frac{{\left(x\right)}^2-{\left({\displaystyle \frac{1}{x}}\right)}^2}{x-\frac{1}{x}} = \frac{\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)}{x-\frac{1}{x}} \\ B = x+\frac{1}{x} = 3^{\frac{1}{4}}+\frac{1}{3^{{\displaystyle \frac{1}{4}}}} = \sqrt[4]{3}+\frac{1}{\sqrt[4]{3}} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} C = \frac{2}{\sqrt{3}\left(\sqrt[4]{3}+{\displaystyle \frac{1}{\sqrt{\sqrt{3}}}}\right)}+\frac{3}{\sqrt[4]{27}} \\ C = \frac{2}{\sqrt{3}\left(\sqrt[4]{3}+{\displaystyle \frac{1}{\sqrt[4]{3}}}\right)}+\frac{3}{\sqrt[4]{27}} \\ C = \frac{2}{\sqrt{3}\left({\displaystyle \frac{\sqrt{3}+1}{\sqrt[4]{3}}}\right)}+\frac{3}{\sqrt[4]{27}} \end{gathered}$$

$$\begin{gathered} C = \frac{2\sqrt[4]{3}}{\sqrt{3}\left({\displaystyle \sqrt{3}+1}\right)}+\frac{3}{\sqrt[4]{27}} \\ C = \frac{2\sqrt[4]{3}}{\sqrt{3}\left({\displaystyle \sqrt{3}+1}\right)}·\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)}+\frac{3}{\sqrt[4]{27}} \\ C = \frac{2\sqrt[4]{3}·\left(\sqrt{3}-1\right)}{\sqrt{3}\left({\displaystyle 3-1}\right)}+\frac{3}{3^{{\displaystyle \frac{3}{4}}}} \\ C = \frac{2·\left(\sqrt{3}-1\right)}{\sqrt[4]{3}·2}+3^{1-\frac{3}{4}} = \frac{\left(\sqrt{3}-1\right)}{\sqrt[4]{3}}+3^{\frac{1}{4}} \end{gathered}$$

$$\begin{gathered} \mathrm{กำหนดให้} 3^{\frac{1}{4}} = y \\ \mathrm{จะได้} C = \frac{\left(y^2-1\right)}{y}+y = \frac{y^2}{y}-\frac{1}{y}+y = y-\frac{1}{y}+y \\ C = 2y-\frac{1}{y} = 2·3^{\frac{1}{4}}-\frac{1}{3^{{\displaystyle \frac{1}{4}}}} = 2·\sqrt[4]{3}-\frac{1}{\sqrt[4]{3}} \\ \mathrm{ดังนั้น} A-B+C \\ = \left(\frac{2}{\sqrt[4]{3}}-\sqrt[4]{3}\right)-\left(\sqrt[4]{3}+\frac{1}{\sqrt[4]{3}}\right)+\left(2·\sqrt[4]{3}-\frac{1}{\sqrt[4]{3}}\right) \\ = 0 \\ \mathbf{ดังนั้น}\mathbf{ }\mathbf{ตอบ}\mathbf{ }\mathbf{5}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{0} \end{gathered}$$