點解我計唔岩展開果d數?
1. (x+3)(2x-5)
我一時計到2x^2-15
一時計到3x^2-15
硬係計唔到2x^2+x-15 哩個正確答案
2. (x+y)^2 - (x-y)^2
我唔知點解會計到0...
但係答案係4xy...
點解全部都計唔岩架=_=
展開到底是什麼東東呀 =_=
點解我計唔岩展開果d數?
2007-08-14 7:32 am
回答 (3)
2007-08-14 8:21 pm
1. (x+3)(2x-5)
=2x^2-5x+6x-15
=2x^2+x-15
2. (x+y)^2 - (x-y)^2
=(x+y)(x+y)-(x-y)(x-y)
=(x^2+xy+xy+y^2)-(x^2-xy-xy+y^2)
=(x^2+2xy+y^2)-(x^2-2xy+y^2)
=x^2+2xy+y^2-x^2+2xy-y^2
=x^2-x^2+2xy+2xy+y^2-y^2
=2xy+2xy
=4xy
=2x^2-5x+6x-15
=2x^2+x-15
2. (x+y)^2 - (x-y)^2
=(x+y)(x+y)-(x-y)(x-y)
=(x^2+xy+xy+y^2)-(x^2-xy-xy+y^2)
=(x^2+2xy+y^2)-(x^2-2xy+y^2)
=x^2+2xy+y^2-x^2+2xy-y^2
=x^2-x^2+2xy+2xy+y^2-y^2
=2xy+2xy
=4xy
參考: me
2007-08-14 7:43 am
1. (x+3)(2x-5)
=x(2x-5)+3(2x-5)
=x(2x)-x(5)+3(2x)-3(5)
=2x^2+x-15
2. (x+y)^2 - (x-y)^2
=(x+y)(x+y)-(x-y)(x-y)
=x(x+y)+y(x+y)-x(x-y)+y(x-y)
=x(x)+x(y)+y(x)+y(y)-x(x)+x(y)+y(x)-y(y)
=4xy
或咁做：
用a^2-b^2=(a+b)(a-b)
2. (x+y)^2 - (x-y)^2
=(x+y+x-y)(x+y-x+y)
=(2x)(2y)
=4xy
唔明再問la
=x(2x-5)+3(2x-5)
=x(2x)-x(5)+3(2x)-3(5)
=2x^2+x-15
2. (x+y)^2 - (x-y)^2
=(x+y)(x+y)-(x-y)(x-y)
=x(x+y)+y(x+y)-x(x-y)+y(x-y)
=x(x)+x(y)+y(x)+y(y)-x(x)+x(y)+y(x)-y(y)
=4xy
或咁做：
用a^2-b^2=(a+b)(a-b)
2. (x+y)^2 - (x-y)^2
=(x+y+x-y)(x+y-x+y)
=(2x)(2y)
=4xy
唔明再問la
參考: own
2007-08-14 7:38 am
第一條
(x+3)(2x-5)
可2睇成
(x+3)(2x) - (x+3)(5)
之後再計
2x^2+6x - (5x+15)
再後就計到
2x^2+x-15
第二條
(x+y)^2 - (x-y)^2 = [(x+y) + (x-y)] [(x+y) - (x-y)]
就會計到4xy
(x+3)(2x-5)
可2睇成
(x+3)(2x) - (x+3)(5)
之後再計
2x^2+6x - (5x+15)
再後就計到
2x^2+x-15
第二條
(x+y)^2 - (x-y)^2 = [(x+y) + (x-y)] [(x+y) - (x-y)]
就會計到4xy
參考: 自己
收錄日期: 2021-04-23 20:22:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070813000051KK06808