因式分解x^2+4xy+4y^2-4(x+2y)+4 要步驟?
回答 (1)
2017-09-09 8:58 am
Sol
x^2+4xy+4y^2－4(x+2y)+4
=(x^2+4xy+4y^2)－4(x+2y)+4
=(x+2y)^2－2*(x+2y)*2+2^2
=(x+2y－2)^2
or
x^2+4xy+4y^2－4(x+2y)+4
=x^2+4xy+4y^2－4x－8y+4
=x^2+(2y)^2+(－2)^2+2*x*(2y)+2*x*(－2)+2*(2y)*(－2)
=(x+2y－2)^2
x^2+4xy+4y^2－4(x+2y)+4
=(x^2+4xy+4y^2)－4(x+2y)+4
=(x+2y)^2－2*(x+2y)*2+2^2
=(x+2y－2)^2
or
x^2+4xy+4y^2－4(x+2y)+4
=x^2+4xy+4y^2－4x－8y+4
=x^2+(2y)^2+(－2)^2+2*x*(2y)+2*x*(－2)+2*(2y)*(－2)
=(x+2y－2)^2
收錄日期: 2021-04-30 22:26:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170909004439AAEKmvn