ข้อสอบคณิต 2 - มีนาคม 2563

$$\begin{gathered} \mathrm{จากโจทย์} \\ \mathrm{พิจารณา} -a \mathrm{กับ} -b \\ \mathrm{จะได้} -a \mathrm{กับ} -b \\ \frac{99}{100} \frac{100}{101} \\ \left(99\right)·\left(101\right) {100}^2 \\ \left(100-1\right)\left(100+1\right) {100}^2 \end{gathered}$$

$$\begin{gathered} \cancel{{100}^2}-1 \cancel{{100}^2} \\ -1 < 0 \\ -a < -b \\ a > b \end{gathered}$$

$$\begin{gathered} \mathrm{พิจารณา} -b \mathrm{กับ} -c \\ \frac{100}{101} \frac{101}{102} \\ 100·102 {101}^2 \\ \left(101-1\right)·\left(101+1\right) 1{01}^2 \\ {101}^2-1 {101}^2 \end{gathered}$$

$$\begin{gathered} -1 < 0 \\ -b < -c \\ b > c \\ \mathrm{ดังนั้น} c < b < a \to \mathrm{ตอบ} 5 \end{gathered}$$