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標題: 兩題 Form 2 數學題 (Rational numbers & Identities) [打印本頁]

作者: anthony 時間: 14/7/2005 04:18 PM
標題: 兩題 Form 2 數學題 (Rational numbers & Identities)
(1) If x^2+4x-1 is identical to (x+p)^2+q, find the values of p and q.
我可否用以下方法來計算呢？
x^2+4x-1 ≡ (x+p)^2+q
x^2+4x+4 ≡ (x+p)^2+q+5
(x+2)^2 ≡ (x+p)^2+q+5
(x+2)^2-5 ≡ (x+p)^2+q

Therefore,
(x+2)^2 = (x+p)^2
and
q = -5
Therefore,
p=2 and q=-5

(2) Rationalize the denominators.
a. 開方6/(2x開方3)
b. 1/(開方2+1)
c. 開方3/(開方2+1)
d. (5x開方2)/(開方3-開方2)

[ Last edited by anthony on 2005-7-23 at 06:45 PM ]

作者: 細薯 時間: 14/7/2005 05:34 PM
第一題用completing square
令到可以變成(x+p)^2+q

x^2+4x-1
=x^2+4x+2^2-5
=(x+2)^2-5
p=2,q=-5

(1) If x^2+4x-1 is identical to (x+p)^2+q, find the values of p and q.
我可否用以下方法來計算呢？
x^2+4x-1 ≡ (x+p)^2+q
x^2+4x+4 ≡ (x+p)^2+q+5
(x+2)^2 ≡ (x+p)^2+q+5
(x+2)^2-5 ≡ (x+p)^2+q

Therefore,
(x+2)^2 = (x+p)^2
and
q = -5
Therefore,
p=2 and q=-5

你這個答案有點問題
這樣會好D↓

x^2+4x-1 ≡ (x+p)^2+q
x^2+4x+4-5 ≡ (x+p)^2+q
(x+2)^2-5 ≡ (x+p)^2+q
(x+2)^2-5 ≡ (x+p)^2+q

第二題

(2) Rationalize the denominators.
a. √6/(2√3)
b. 1/(√2+1)
c. √3/(√2+1)
d. (5√2)/(√3-√2)

a.
√6/(2√3)
=√(2x3)  /(2√3)
=√2√3  /(2√3)
=√2/2

b.
1/ (√2+1)
=(√2-1) /  (√2+1)(√2-1)
=(√2-1)/1
=√2-1

c.
√3/(√2+1)
=√3(√2-1) / (√2+1)(√2-1)
=√3(√2-1) / 1
=√3(√2-1)

d. (5√2)/(√3-√2)
=(5√2)(√3+√2) / (√3-√2)(√3+√2)
=5√6 + 10 /1
=5√6 + 10

有冇錯．．﹖

作者: anthony 時間: 14/7/2005 06:37 PM
Thank you very much. [E-Hap]

作者: 抹茶可樂 時間: 15/7/2005 12:11 AM
第一條你用的方法屬「高技巧」
直接拆開就得了~XD

(x+2)^2-5 ≡ (x+p)^2+q
做到呢到時可直接說
p=2
q=-5

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