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发表于 29-11-2010 10:19 PM
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我有用过completing the square
2+2x-x^2
y=-x^2+4x+4
=-[(x-2)^2-8]
答案是maximum point y is 4
...
hongji 发表于 29-11-2010 10:13 PM 
不对不对。。。
f(x) = 2+2x-[x^2/2]
f(x)=(-1/2)[x^2-4x]+2
f(x)=(-1/2)[(x-2)^2-4]+2
f(x)=(-1/2)(x-2)^2+4 |
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发表于 30-11-2010 07:44 AM
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不对不对。。。
f(x) = 2+2x-[x^2/2]
f(x)=(-1/2)[x^2-4x]+2
f(x)=(-1/2)[(x-2)^2-4]+2
...
Allmaths 发表于 29-11-2010 10:19 PM 
阿。。原来是这样。感激万分阿。。
不过还有。。我不明白怎样找出来
if @ is a root of the equation x^2 + 3x +1 =0 ,show that @^2=-3@-1 and @^3=-3@^2-@.Deduce that @^3=8@+3.If β is the other root of the equation,hence find the value of @^3+β^3. |
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发表于 30-11-2010 07:48 AM
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if [x-1/x=p] ,express [x^2+1/x^2] in the terms of p.By using the method of substitution,solve the equation [6x^4-25x^3-12^2+25x+6=0]
我会addition不会substitution.. |
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发表于 30-11-2010 09:58 AM
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阿。。原来是这样。感激万分阿。。
不过还有。。我不明白怎样找出来
if @ is a root of th ...
hongji 发表于 30-11-2010 07:44 AM 
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发表于 30-11-2010 10:16 AM
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if [x-1/x=p] ,express [x^2+1/x^2] in the terms of p.By using the method of substitution,solve the eq ...
hongji 发表于 30-11-2010 07:48 AM 
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发表于 1-12-2010 07:04 PM
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Allmaths 发表于 30-11-2010 10:16 AM 
好厉害哦谢谢。。 |
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发表于 1-12-2010 07:12 PM
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我做了一个晚上
答案还不懂怎样找
express each of the following in partial fractions
[9-8x]/[(2x-1)(3-x)]
Hence,prove that
[(9-8x)^2]/[(2x-1)^2(3-x)^2] = 4/[(2x-1)^2]+9/[(3-x)^2-24/[5(2x-1)]-12/[5(3-x)
=A/[2x-1]+B/[2x-1]^2+C/[3-x]+D/[3-x]^2
我找到B和D
找不到A and C |
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发表于 1-12-2010 07:16 PM
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y^2+[9/y^2]-4(y+3/y)-6=0
这题。。用formula:x+1/x的做不到。。
还有其他formula的吗? |
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发表于 1-12-2010 07:29 PM
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y^2+[9/y^2]-4(y+3/y)-6=0
这题。。用formula:x+1/x的做不到。。
还有其他formula的吗?
hongji 发表于 1-12-2010 07:16 PM 
y^2+[9/y^2]-4(y+3/y)-6 =y^2+[3^2/y^2]-4(y+3/y)-6
= (y+3/y)^2 - 2y(3/y) -4(y+3/y)-6
let a = y + 3/y
然后自己来 |
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发表于 1-12-2010 07:37 PM
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回复 2247# hongji
[9-8x]/[(2x-1)(3-x)] = a/(2x-1) + b/(3-x)
=a(3-x) + b(2x-1) / (2x-1)(3-x)
a(3-x) + b(2x-1) = 9-8x
x=3 , 5b = -15
b=-3
x=1/2 , 5a/2 = 5
a=2
[9-8x]/[(2x-1)(3-x)] = 2/(2x-1) - 3/(3-x)
------------
[9-8x]^2/[(2x-1)(3-x)]^2 = [ 2/(2x-1) - 3/(3-x)] ^2
= 4/(2x-1)^2 + 9/(3-x)^2 - 2 [2/(2x-1)][3/(3-x)]
= 4/(2x-1)^2 + 9/(3-x)^2 -12/[(2x-1)(3-x)]
12/[(2x-1)(3-x)] = a/(2x-1) + b/(3-x)
=a(3-x) + b(2x-1) / (2x-1)(3-x)
a(3-x) + b(2x-1) =12
x=3 , 5b=12
b=12/5
x=1/2 , 5a/2=12
a=24/5
12/[(2x-1)(3-x)] = 24/5(2x-1) + 12/5(3-x)
[9-8x]^2/[(2x-1)(3-x)]^2 = 4/(2x-1)^2 + 9/(3-x)^2 -[24/5(2x-1) + 12/5(3-x)]
= 4/(2x-1)^2 + 9/(3-x)^2 -24/5(2x-1) - 12/5(3-x) |
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发表于 1-12-2010 07:44 PM
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y^2+[9/y^2]-4(y+3/y)-6 =y^2+[3^2/y^2]-4(y+3/y)-6
= (y+3/y)^2 ...
peaceboy 发表于 1-12-2010 07:29 PM 
那个。。(y+3/y)^2-2y(3/y)...怎样分得啊? |
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发表于 1-12-2010 07:46 PM
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那个。。(y+3/y)^2-2y(3/y)...怎样分得啊?
hongji 发表于 1-12-2010 07:44 PM 
y^2 + (3/y)^2
let y=a , 3/y =b
a^2+b^2 = (a+b)^2 -2ab
then , y^2 + (3/y)^2 = (y+3/y)^2-2y(3/y) |
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发表于 1-12-2010 07:53 PM
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回复 hongji
[9-8x]/[(2x-1)(3-x)] = a/(2x-1) + b/(3-x)
=a ...
peaceboy 发表于 1-12-2010 07:37 PM 
好快哦。。我抄完了
谢谢。。 |
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发表于 1-12-2010 07:56 PM
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y^2 + (3/y)^2
let y=a , 3/y =b
a^2+b^2 = (a+b)^2 -2ab
then , y^2 + (3/y)^2 = (y+3 ...
peaceboy 发表于 1-12-2010 07:46 PM 
我会了。。谢谢 |
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发表于 1-12-2010 08:02 PM
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还有我还不是很会分
x^4 + 5x^3 -2x^2-5x+1=0
divide by x^2
x^2+[5x]-2-[5/x]+[1/x^2]=0
x^2+[1/x^2]+5(x-1/x)-2=0
那么。。x^2+1/x^2=u^2-2
x-1/x=-u^2? |
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发表于 1-12-2010 08:44 PM
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回复 2255# hongji
a^2+b^2 = (a-b)^2 +2ab |
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发表于 2-12-2010 07:51 AM
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1)if @ and β are the roots of the quadratic equation x^2-mx+n=0 , show that (@-m)^2 +(β-m)^2=m^2-2n
2)given @ and β are the roots of the quadratic equation px^2+qx+r=0 find the relationship between p , q and r if
a)@=2β+1 b)@=3β+1
3)if @ and β are the roots of the quadratic equation x^2+ax+a^2+b^2=0,show that @^4+[@^2β^2]+β4=b^2(2a^2+3b^2)
4)the quadratic equation x^2+ax +b=0 and x^2+bx+a=0 have a common root .Show that either a=b or a+b=1 |
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发表于 2-12-2010 10:27 AM
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1)if @ and β are the roots of the quadratic equation x^2-mx+n=0 , show that (@-m)^2 +(β-m)^2=m^2-2n
hongji 发表于 2-12-2010 07:51 AM 
(1)(@-m)^2+(β-m)^2=@^2+β^2-2m(@+β)+2m^2
=(@+β)^2-2@β-2m(@+β)+2m^2
=m^2-2n-2m^2+2m^2
=m^2-2n (shown) |
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发表于 2-12-2010 10:33 AM
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3)if @ and β are the roots of the quadratic equation x^2+ax+a^2+b^2=0,show that @^4+[@^2β^2]+β^4=b^2(2a^2+3b^2)
hongji 发表于 2-12-2010 07:51 AM 
(3)@^4+[@^2β^2]+β^4=(@^2+β^2-@β)(@^2+β^2+@β)
=[(@+β)^2-3@β][(@+β)^2-@β]
=[(-a)^2-3(a^2+b^2)][(-a)^2-(a^2+b^2)]
=b^2(2a^2+3b^2) (shown) |
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发表于 2-12-2010 11:19 AM
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4)the quadratic equation x^2+ax +b=0 and x^2+bx+a=0 have a common root .Show that either a=b or a+b=1
hongji 发表于 2-12-2010 07:51 AM 
是a+b=-1吗? |
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