ข้อสอบคณิต O-NET - 2551

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{สามารถกระจายสูตรกำลังสองสมบูรณ์ได้} \\ \mathrm{จะได้} {\left(a+\frac{1}{\left|a\right|}\right)}^2 = a^2+2·\frac{a}{\left|a\right|}+{\left(\frac{1}{\left|a\right|}\right)}^2 \\ = a^2+2·\frac{a}{\left|a\right|}+\frac{1}{{\left|a\right|}^2} \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} {\left(\left|a\right|-\frac{1}{a}\right)}^2 = {\left|a\right|}^2-2·\left|a\right|·\frac{1}{a}+{\left(\frac{1}{a}\right)}^2 \\ = a^2-2·\frac{\left|a\right|}{a}+\frac{1}{a^2} \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \overline{)1} - \overline{)2} : x = 2·\frac{a}{\left|a\right|}+2·\frac{\left|a\right|}{a} \\ \mathrm{จะได้} \frac{a}{\left|a\right|} = \left\{\begin{array}{ll}1& ; a > 0\\ -1& ; a < 0\end{array}\right. \\ \frac{\left|a\right|}{a} = \left\{\begin{array}{ll}1& ; a > 0\\ -1& ; a < 0\end{array}\right. \\ \mathrm{ดังนั้น} x = 2+2 = 4 \end{gathered}$$