ข้อสอบคณิต O-NET - 2550

$$\begin{gathered} \mathrm{จากตัวเลือก} \\ 1) \mathrm{จาก} \sqrt{0.9+10} < \sqrt{0.9}+\sqrt{10} \\ \mathrm{กำลัง} 2 \mathrm{ทั้ง} 2 \mathrm{ข้าง} \mathrm{จะได้} \\ 0.9+10 < 0.9+2\sqrt{0.9}\sqrt{10}+10 \\ 10.9 < 10.9+2\sqrt{0.9}\sqrt{10} \\ 0 < 2\sqrt{0.9}\sqrt{10} \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} 2) \mathrm{จาก} {\left(0.9\right)}^{\frac{1}{2}}·{\left(0.9\right)}^{\frac{1}{4}} < 0.9 \\ \mathrm{จะได้} {\left(0.9\right)}^{\frac{1}{2}+\frac{1}{4}} < 0.9 \\ {\left(0.9\right)}^{\frac{3}{4}} < {\left(0.9\right)}^1 \\ \mathrm{จำนวนที่น้อยกว่า} 1 \mathrm{ยิ่งยกกำลังมาก} \mathrm{จะยิ่งได้ค่าน้อย} \\ \mathrm{ดังนั้น} {\left(0.9\right)}^{\frac{3}{4}} < {\left(0.9\right)}^1 \mathrm{ผิด} \end{gathered}$$

$$\begin{gathered} 3) \mathrm{จาก} {\left(0.9\right)}^{\frac{1}{2}}·{\left(1.1\right)}^{\frac{1}{3}} < {\left(1.1\right)}^{\frac{1}{2}}·{\left(0.9\right)}^{\frac{1}{3}} \\ \mathrm{จะได้} \frac{{\left(0.9\right)}^{{\displaystyle \frac{1}{2}}}}{{\left(0.9\right)}^{\frac{1}{3}}} < \frac{{\left(1.1\right)}^{\frac{1}{2}} }{{\left(1.1\right)}^{\frac{1}{3}} } \\ {\left(0.9\right)}^{\frac{1}{2}-\frac{1}{3}} < {\left(1.1\right)}^{\frac{1}{2}-\frac{1}{3}} \\ {\left(0.9\right)}^{\frac{1}{6}} < {\left(1.1\right)}^{\frac{1}{6}} \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} 4) \mathrm{จาก} {\left(125\right)}^{\frac{1}{300}} < {\left(100\right)}^{\frac{1}{200}} \\ \mathrm{จะได้} {\left({125}^{\frac{1}{3}}\right)}^{\frac{1}{100}} < {\left({100}^{\frac{1}{2}}\right)}^{\frac{1}{100}} \\ 5^{\frac{1}{100}} < {10}^{\frac{1}{100}} \mathrm{ถูก} \end{gathered}$$