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发表于 24-7-2009 10:42 PM
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回复 360# 末日审判 的帖子
有答案吗?怕做错~ |
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楼主 |
发表于 24-7-2009 10:56 PM
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原帖由 末日审判 于 24-7-2009 09:15 PM 发表
帮帮我下,
1.The sum of the first four terms of a G.P is five times the sum of the sum of the first two terms.Find the value of the common ratio,r,given that r>1.
2.The 1st term of a G.P. exceeds ...
我想看高手做第一题
2. a-ar = 1
a(1-r)=1
a=1/1-r ---1
S3=13/5
a(1-r^3)/1-r =13/5 ----2
substitute 1 into 2
(1-r)(1-r^3)/1-r =13/5
然后你做玩后,会有一个cubic equation出来,然后你factorise 后就是答案了.
r=2/5 ;a=5/3 or
r= -4 ; a=1/5 |
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发表于 24-7-2009 11:00 PM
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发表于 24-7-2009 11:02 PM
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回复 360# 末日审判 的帖子
a)let GP =a,ar,ar^2,and sum of the first n term of series=Sn
given that S4=5S2
a(1-r^4)/1-r=5a(1-r^2)/1-r
1-r^4=5(1-r^2)
(1+r^2)(1-r^2)=5(1-r^2)
1+r^2=5
r^2=4
r=2 |
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发表于 24-7-2009 11:07 PM
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Sum of 1st 4T = a(r^4 - 1)/(r - 1) ...........................A
Sum of 1st 2T = a(r^2 - 1)/(r - 1) ...........................B
A = 5B
(r^4 - 1) = 5(r^2 - 1)
r^4 - 5r^2 + 4 = 0
let r^2 =x
x^2 - 5x + 4= 0
x = 1, 4
then , r = 1, -1, 2, -2
since r >1, then r = 2 |
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发表于 24-7-2009 11:08 PM
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发表于 24-7-2009 11:09 PM
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回复 365# 四月一日的小皮 的帖子
same answer |
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发表于 25-7-2009 09:59 AM
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原帖由 白羊座aries 于 24-7-2009 10:56 PM 发表
我想看高手做第一题
2. a-ar = 1
a(1-r)=1
a=1/1-r ---1
S3=13/5
a(1-r^3)/1-r =13/5 ----2
substitute 1 into 2
(1-r)(1-r^3)/1-r =13/5
然后你做玩后,会有一个cubic equation出来,然后你 ... 原帖由 ynnee 于 24-7-2009 11:02 PM 发表
a)let GP =a,ar,ar^2,and sum of the first n term of series=Sn
given that S4=5S2
a(1-r^4)/1-r=5a(1-r^2)/1-r
1-r^4=5(1-r^2)
(1+r^2)(1-r^2)=5(1-r^2)
1+r^2=5
... 原帖由 四月一日的小皮 于 24-7-2009 11:07 PM 发表
Sum of 1st 4T = a(r^4 - 1)/(r - 1) ...........................A
Sum of 1st 2T = a(r^2 - 1)/(r - 1) ...........................B
A = 5B
(r^4 - 1) = 5(r^2 - 1)
r^4 - 5r^2 + 4 = 0
let r^ ...
你们答案都对了,很谢谢你们!
你们用来解决问题的方式都不同,我也学到了!
谢谢! |
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发表于 26-7-2009 02:55 PM
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我又有新問題了,希望大家能幫忙!
1.The first three of four integers are in A.P.and the last three are in a G.P.Find these four numbers given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.(answer:12,16,20,25)
2.The 2nd,6th and 8th terms of an A.P.are three successive terms of a G.P.Find the common ratio of the G.P.and obtain an expression for the nth term of the G.P.(answer:r=1/2;16/9 a(1/2)^n
3.Prove that 3^x,3^y,and3^z are successive terms of a G.P.,if x,y,and z are successive terms of an A.P. |
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发表于 26-7-2009 09:56 PM
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发表于 26-7-2009 10:03 PM
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from the ap, the 3 following gp is shown :
a+d , a+ 5d ,a +7d
since common ratio is same,then
(a+5d)/(a+d) = (a+ 7d)/(a+5d)
(a+5d)^2 = (a+7d)(a +d)
a^2 + 10ad + 25d^2 = a^2 + 8ad + 7d^2
2ad = - 18 d^2
a = -9 d
thus, r = ( -9d + 5d) / (-9d + d)
= -4d / -8d = 1/2
the gp can form the following series using d= -a/9
(8/9)a,(4/9)a,(2/9)a.....
1st term = 8a/9
eqn -----> term n = (8/9)a (1/2)^(n-1)
= (8/9)(1/2)a(1/2)^n
= 16/9 a(1/2)^n
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发表于 26-7-2009 10:06 PM
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x,y,z is ap thus
y - x = z - y
let m = y -x = z - y
then, 3^m = 3^(y - x) = 3 ^(z -y)
3^y / 3^x = 3^z / 3^y.............> common ratio
thus 3^x , 3^y , 3^z is a gp series.
(correct me if i was wrong.... ) |
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发表于 26-7-2009 11:17 PM
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2a + 3d = 36
d = 12 - 3a/2 ..............................1
d不是等于d=12-2a/3吗?
term 1 + term 4 = 37
.........
(i simplify the eqn to become this)
a(a + d) + (a + d)^2 = 37 (a + d).....................................................2
subtitute eqn 1 into 2 :
.....(again, i simplify it since it too complex to write out )
.......
这边不明白,可否详细点? |
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发表于 27-7-2009 05:42 PM
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原帖由 末日审判 于 26-7-2009 11:17 PM 发表
d不是等于吗?
这边不明白,可否详细点?
ummm.....yes, is d=12-2a/3...mistake.
term 1 = a
term 4 = (a + 2d)^2 / (a + d)
then...
a + (a + 2d)^2 / (a + d) = 37
(multiply both side with (a+d):
a(a+d) + (a+ 2d)^2 = 37(a+d) |
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发表于 6-8-2009 06:41 PM
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∫ ln x dx 該怎麽solve?
:@
我算到要吐血了.... |
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发表于 6-8-2009 08:54 PM
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原帖由 Dicardo 于 6-8-2009 06:41 PM 发表
∫ ln x dx 該怎麽solve?
:@
我算到要吐血了....
你可以用Integration by part.
我额外补充一下。当你们用integration by part时,
你们得知道一些 rule,就是选出 first function当成你们的fixed function.
例如:
∫ f(x). g(x) dx = f(x) ∫ g(x) dx - ∫ [ ∫ g(x) dx ] . f '(x) dx
In words, this formula states that
The integral of the product of two functions
= first function x integral of second - integral of ( integral of second x diff. coefficient of first )
所以你们先在两个functions中锁定其中一个function为你的first function(我习惯用Fixed这个字眼).
其实,这里有个我个人的口诀。
By using integration by part,
Fixed . Integrate - ∫ Integrate . differentiate Fixed dx
接下来,我教教你们如何 辨认 哪个 functions 为 你们的 first function.
这里就是要obey the rules 来找出 first function。
The first function is chosen as the function which comes first in the word ILATE, where
I - stands for Inverse function (Exp: f(x) = sin^(-1) x; cos^(-1) )
L - stands for Logarithmic function (Exp: f(x) = log x; ln x )
A - stands for Algebraic function (Exp: f(x) = 2x+3 )
T - stands for Trigonometric function (Exp: f(x) = sin x; cos x )
E - stands for Exponential function (Exp: f(x) = e^(x) )
通常STPM的level, 只有以下的rules:
你们根据以上的rules就可以找出fixed 就是 (first function)。
回你的问题:
∫ ln x dx
这个问题有两个functions就是
∫ ln x dx
= ∫ 1 . ln x dx
设 f(x) = ln x 和 g(x) = 1
你现在可以很清楚地 辨认 哪一个是 fixed 吗?
就是用刚才我所说的rules来找出fixed.
其实 ln x 是 属于 Logarithmic function,
而 1 是 属于 Algebraic function
根据rules:
你看到lnx 排在 x 的上面,所以 lnx 就是 设为 Fixed.
然后根据这个
∫ f(x). g(x) dx = f(x) ∫ g(x) dx - ∫ [ ∫ g(x) dx ] . f '(x) dx
或
∫ f(x). g(x) dx = Fixed . Integrate - ∫ Integrate . differentiate Fixed dx
Solution:
设
f(x) = ln x 为 Fixed
和
g(x) = 1 为 second
∫ ln x dx
= ∫ 1 . ln x dx
= ln x . x - ∫ x . 1/ x dx {因为 Fixed . Integrate - ∫ Integrate . differentiate Fixed dx }
= x.lnx - ∫ 1 dx
= x.lnx - x + c
= x ln (x/e) + c
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楼主 |
发表于 10-8-2009 10:59 PM
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by substituting y= x + 1/x
f(x)= x^3 -4x -6 -4/x +1/(x^3)
y^2-2=x^2+1/x^2
然后要怎样 |
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发表于 10-8-2009 11:43 PM
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原帖由 白羊座aries 于 10-8-2009 10:59 PM 发表
by substituting y= x + 1/x
f(x)= x^3 -4x -6 -4/x +1/(x^3)
y^2-2=x^2+1/x^2
然后要怎样
f(x) = x^3 + 1/(x^3) - 4x - 4/x - 6
= x^2 + 1/x^2 - 4 - 4/x^2 - 6/x (divide x)
=(y^2 - 2) - .............(我忘记了) |
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发表于 11-8-2009 12:59 AM
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原帖由 白羊座aries 于 10-8-2009 10:59 PM 发表
by substituting y= x + 1/x
f(x)= x^3 -4x -6 -4/x +1/(x^3)
y^2-2=x^2+1/x^2
然后要怎样
f(x) = x^3 - 4x - 6 - 4/x + 1/(x^3)
f(x) = x^3 + 1/(x^3) - 4x - 4/x -6
since y= x + 1/x
=> y^3 = (x + 1/x)^3
=> y^3 = [ x^3 + 1/(x^3) ] + 3(x + 1/x)
=> y^3 = [ x^3 + 1/(x^3) ] + 3y
=> y^3 -3y = [ x^3 + 1(/x^3) ]
Therefore,
f(x) = x^3 + 1/(x^3) - 4x - 4/x -6
f(x) =[ x^3 + 1/(x^3) ] - 4(x + 1/x) -6
f(y) = [y^3 -3y] - 4y -6
f(y) = y^3 -7y - 6 |
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发表于 11-8-2009 01:03 AM
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