ข้อสอบคณิต 9 สามัญ - ปี 2563

$$\begin{gathered} \mathrm{จากโจทย์} \\ 12\left(4^x\right)+18\left(9^x\right) = 35\left(6^x\right) \\ 12{\left({\left(2\right)}^2\right)}^x+18{\left(3^2\right)}^x = 35{\left(2·3\right)}^x \\ 12\left(2^{2x}\right)-35\left(2^x\right)\left(3^x\right)+18\left(3^{2x}\right) = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{ให้} 2^x = a \mathrm{และ} 3^x= b \\ 12a^2-35ab+18b^2 = 0 \\ \left(3a-2b\right)\left(4a-9b\right) = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} 3a = 2b \\ 3\left(2^x\right) = 2\left(3^x\right) \\ 2^{x-1} = 3^{x-1} \\ x-1 = 0 \\ x = 1 \end{gathered}$$

$$\begin{gathered} \mathrm{หรือ} \\ 4a = 9b \\ 4\left(2^x\right) = 9\left(3^x\right) \\ 2^{x+2} = 3^{x+2} \\ x+2 = 0 \\ x = -2 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \mathrm{ผลบวกคำตอบ} = 1+\left(-2\right) = -1 \\ \mathrm{ตอบ} 1 \end{gathered}$$