有2條數:
第一條話計出答案係咩
(4x+y)^2 - (4x-y)^2
之前教中間就咁跳曬step寫
(4x+y+4x-y) + (4x+y-4x+y)<--呢條點嚟...
=8x(2y) = 16xy-_-
第2條:
if p & q are constant no. and x^2+p = (x+2)(x+q)+10, what is p?
有2條數唔識急求解答
2013-04-24 9:31 pm
回答 (1)
2013-04-24 11:22 pm
✔ 最佳答案
1.由於 u² - v² = (u + v)(u - v)
故此 (4x + y)² - (4x - y)²
= [(4x + y) + (4x - y)] [(4x + y) - (4x - y)]
= (4x + y + 4x - y)(4x + y - 4x + y)
= (8x)(2y)
= 16xy
2.
x² + p = (x + 2) (x + q) + 10
x² + p = (x² + qx + 2x + 2q) + 10
x² + p = x² + (q + 2)x + (2q + 10)
兩邊 x 項的係數相等：
q + 2 = 0
q = -2
兩邊常數項相等：
p = 2q + 10
p = 2(-2) + 10
p = 6
參考: fooks
收錄日期: 2021-04-13 19:26:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130424000051KK00109