併項一條
x²-4y²+x+2y
列式計算
thx
因式分解..............................
2007-01-19 2:33 am
回答 (9)
2007-01-19 2:36 am
✔ 最佳答案
x²-4y²+x+2y=(x-2y)(x+2y)+(x+2y)
=(x+2y)(x-2y+1)
2007-01-20 5:10 pm
=(x²-4y²)+(x+2y)
=[x²-(2y²)]+(x+2y)
=(x+2y)(x-2y)+(x+2y)
=(x+2y)[(x-2y)+1]
=(x+2y)(x-2y+1)
=[x²-(2y²)]+(x+2y)
=(x+2y)(x-2y)+(x+2y)
=(x+2y)[(x-2y)+1]
=(x+2y)(x-2y+1)
2007-01-19 4:23 am
x²-4y²+x+2y
=x(x+1)-2y(2y-1)
=x(x+1)-2y(2y-1)
2007-01-19 3:46 am
x²-4y²+x+2y
=x²-(2y)²+x+2y
=(x+2y)(x-2y)+(x+2y)
=(x+2y)(x-2y+1)
=x²-(2y)²+x+2y
=(x+2y)(x-2y)+(x+2y)
=(x+2y)(x-2y+1)
參考: my brother
2007-01-19 2:40 am
首先將同樣的放在一起 : x²-4y²+x+2y → x² +x-4y²+2y
之後就可以抽因子,留意正負號 : x(x+1)-2y(2y-1)
這就是答案......
之後就可以抽因子,留意正負號 : x(x+1)-2y(2y-1)
這就是答案......
參考: 自己
2007-01-19 2:36 am
x²-4y²+x+2y
=x
=x
收錄日期: 2021-04-12 17:53:26
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