x²-2xy+y²-4x²+4y+3要因式分解 各位大大幫幫忙~~?
回答 (2)
2015-11-25 10:51 am
✔ 最佳答案
x²-2xy+y²-4x²+4y+3= (x-y)²-(2x)²+4y+3
= (x-y-2x)(x-y+2x)+4y+3
= (-x-y)(3x-y)+4y+3
設-x-y=a,3x-y=b, -(3a+b)=4y
= ab-(3a+b)+3
= ab-3a-(b-3)
= a(b-3)-(b-3)
= (a-1)(b-3)
= (-x-y-1)(3x-y-3)
2015-12-24 4:20 am
x²-2xy+y²-4x²+4y+3
= (x-y)²-(2x)²+4y+3
= (x-y-2x)(x-y+2x)+4y+3
= (-x-y)(3x-y)+4y+3
設-x-y=a,3x-y=b, -(3a+b)=4y
= ab-(3a+b)+3
= ab-3a-(b-3)
= a(b-3)-(b-3)
= (a-1)(b-3)
= (-x-y-1)(3x-y-3)
= (x-y)²-(2x)²+4y+3
= (x-y-2x)(x-y+2x)+4y+3
= (-x-y)(3x-y)+4y+3
設-x-y=a,3x-y=b, -(3a+b)=4y
= ab-(3a+b)+3
= ab-3a-(b-3)
= a(b-3)-(b-3)
= (a-1)(b-3)
= (-x-y-1)(3x-y-3)
收錄日期: 2021-04-18 14:05:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151125013828AA7vP4w