ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathbf{พิจารณา}\mathbf{ }\mathbf{a} \mathrm{a}=\sqrt{7+4\sqrt{3}} \\ =\sqrt{7+2\sqrt{12}} \\ \mathrm{จาก} \sqrt{\mathrm{a}+\mathrm{b}+2\sqrt{\mathrm{ab}}}=\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}} \\ \mathrm{จะได้} \mathrm{a}=\sqrt{4}+\sqrt{3} \\ =2+\sqrt{3} \\ =2+1.732 \\ =3.732 \end{gathered}$$

$$\begin{gathered} \mathbf{พิจารณา}\mathbf{ }\mathit{b} \mathrm{b}=\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}} \\ \mathrm{ยกกำลัง} 2 \mathrm{ทั้งสองข้าง} ; \\ {\mathrm{b}}^2=2\sqrt{2\sqrt{2\sqrt{2...}}} \\ {\mathrm{b}}^2=2\mathrm{b} \\ {\mathrm{b}}^2-2\mathrm{b}=0 \\ \mathrm{b}\left(\mathrm{b}-2\right)=0 \\ \mathrm{b}=0,2 ; \mathrm{โดย} \mathrm{b}\ne 0 \\ \mathrm{b}=2 \end{gathered}$$

$$\begin{gathered} \mathbf{พิจารณา}\mathbf{ }\mathbf{c} \mathrm{c}=\sqrt{2}+\sqrt{3} \\ =1.418+1.732 \\ =3.146 \\ \mathrm{ดังนั้น} \mathrm{b}<\mathrm{c}<\mathrm{a} \\ \frac{1}{\mathrm{b}}>\frac{1}{\mathrm{c}}>\frac{1}{\mathrm{a}} \end{gathered}$$