3x²-2x-4=0，則（x-2)(3x+4)(x+1)(3x-5)=？ （（副算式，謝謝?
回答 (2)
2019-01-17 12:36 am
3x² -2x = 4
(x - 2)(3x + 4)(x + 1)(3x - 5)
= (3x² - 2x - 8)(3x² - 2x - 5)
= (4 - 8)(4 - 5)
= 4 #
(x - 2)(3x + 4)(x + 1)(3x - 5)
= (3x² - 2x - 8)(3x² - 2x - 5)
= (4 - 8)(4 - 5)
= 4 #
2019-01-12 3:49 pm
3x^2－2x－4=0，則(x－2)(3x+4)(x+1)(3x－5)=？
Sol
3x^2－2x－4=0
3x^2－2x=4
(x－2)(3x+4)(x+1)(3x－5)
=[(x－2)(3x+4)]*[(x+1)(3x－5)]
=(3x^2－2x－8)(3x^2－2x－5)
=(4－8)*(4－5)
=4
Sol
3x^2－2x－4=0
3x^2－2x=4
(x－2)(3x+4)(x+1)(3x－5)
=[(x－2)(3x+4)]*[(x+1)(3x－5)]
=(3x^2－2x－8)(3x^2－2x－5)
=(4－8)*(4－5)
=4
收錄日期: 2021-04-30 22:48:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190112072915AAqRZbo