大家好^_^想問問（x^2+2x+1),(9x^2-6x+1)的比例中項=？謝謝！🙇🏻♀️?
回答 (1)
2021-01-16 5:46 pm
設比例中項=f(x)
所以f(x)/（x^2+2x+1) = (9x^2-6x+1)/f(x)
-->[f(x)]^2 = （x^2+2x+1)(9x^2-6x+1) = (x+1)^2(3x-1)^2 = [(x+1)(3x-1)]^2
所以比例中項=f(x)= (x+1)(3x-1) = 3x^2+2x-1 ..ans
所以f(x)/（x^2+2x+1) = (9x^2-6x+1)/f(x)
-->[f(x)]^2 = （x^2+2x+1)(9x^2-6x+1) = (x+1)^2(3x-1)^2 = [(x+1)(3x-1)]^2
所以比例中項=f(x)= (x+1)(3x-1) = 3x^2+2x-1 ..ans
收錄日期: 2021-04-11 23:25:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20210116094150AAXCW9k