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neokeynesian
10-29-2004, 06:08 AM
Please see the attachment

For question 2,

(2.sqrt(ab))^2 = 2.ab not 4.ab??
anyone can explains?

thanks

lmtuan
10-29-2004, 07:06 AM
question 1
[1 / sqrt(5) ] * { [1 + sqrt(5)]/2 - [1 - sqrt(5)]/2 }
= [1 / sqrt(5)] * { [1 + sqrt(5) -1 + sqrt(5)] / 2}
= [1 / sqrt(5)] * { [2 * sqrt(5)] / 2}
= [1 / sqrt(5)] * sqrt(5)
= 1
Thus, answer is B

question 2:
Note that a>0 and b>0 , then sqrt(ab) = sqrt(a)*sqrt(b), sqrt(a)>0, sqrt(b)>0, and [sqrt(a) + sqrt(b)]>0.
We have:
sqrt{a + b + 2*sqrt(ab)}
= sqrt{ [sqrt(a)]^2 + 2*sqrt(a)*sqrt(b) + [sqrt(b)]^2 }
= sqrt{ [sqrt(a) + sqrt(b)]^2 }
= abs{sqrt(a) + sqrt(b)}
= sqrt(a) + sqrt(b).
Thus, the answer of this question is C

lmtuan
10-29-2004, 07:07 AM
Oh, I am sorry. I has a typo.
question 1
[1 / sqrt(5) ] * { [1 + sqrt(5)]/2 - [1 - sqrt(5)]/2 }
= [1 / sqrt(5)] * { [1 + sqrt(5) -1 + sqrt(5)] / 2}
= [1 / sqrt(5)] * { [2 * sqrt(5)] / 2}
= [1 / sqrt(5)] * sqrt(5)
= 1
Thus, answer is C

question 2:
Note that a>0 and b>0 , then sqrt(ab) = sqrt(a)*sqrt(b), sqrt(a)>0, sqrt(b)>0, and [sqrt(a) + sqrt(b)]>0.
We have:
sqrt{a + b + 2*sqrt(ab)}
= sqrt{ [sqrt(a)]^2 + 2*sqrt(a)*sqrt(b) + [sqrt(b)]^2 }
= sqrt{ [sqrt(a) + sqrt(b)]^2 }
= abs{sqrt(a) + sqrt(b)}
= sqrt(a) + sqrt(b).
Thus, the answer of this question is C

vn_snoopy
10-30-2004, 02:22 AM
Just apply (x+y)^2 = x^2 + y^2 + 2xy with x = sqrt(a) > 0 and y = sqrt(b) > 0. ;)

In this case, since x and y are both positive (hence the sum x + y), we can take the square root of (x + y)^2 as (x + y). And... so on ... :)

calm_J
11-06-2004, 11:04 AM
hi guys ,
the complete answer

sabna
11-08-2004, 11:53 AM
wow..nice
appreciable..!!!!!!

prayers
sabna

777
11-08-2004, 08:14 PM
1) When I did the subtraction, I got--

(2sqrt5/2)1/sqrt5 = 2/2 = 1

2) Square both sides and they're equal