107建國中學二招

2
方程式\((4x+1)(6x+1)(8x+1)(24x+1)=21\)之實數解為　　　。
[解答]
\( \displaystyle (4x+1)(6x+1)(8x+1)(24x+1)=21 \)
\( \displaystyle (48x^2+14x+1)(96x^2+28x+1)=21 \)
令 \( \displaystyle A=48x^2+14x \)
\( \displaystyle (A+1)(2A+1)=21 \)