数学Paper 1讨论专区

hongji

回复  hongji


     positive power用一般方式expend有什么问题？

[(2+x-4x^2)]^2 = (2+ x-4x^ ...
peaceboy 发表于 13-12-2010 05:41 PM

    yup x^4=0

hongji

x^2-4xy=28   and   2xy+y^2=-3

how to find x and y?
-4xy=28-x^2
x=4x^2-7/y用这个方法来做出现x^2 ,Y^2两个在一起可以找到吗？

whyyie

回复 2442# hongji

x^2-4xy=28   ...(1)
y^2+2xy=-3  ...(2)

(2)/(1),
y^2+2xy=-3
_______=_
x^2-4xy  28

rearrange得
28y^2 +56xy=12xy-3x^2
3x^2 +44xy+28y^2=0
(3x+2y)(x+14y)=0
y= -2x/3  or x =-14y

Substitute y=-2x/3 into (1),
...
...
Substitute x=-14y into (1),
...
...

peaceboy

回复 2442# hongji


    x^2-4xy=28   and   2xy+y^2=-3

    x^2-4xy=28
4xy = x^2 -28
y = [(x^2-28)/4x]

2x[(x^2-28)/4x] + [(x^2-28)/4x]^2 = -3

(x^2-28)/2 + (x^2-28)^2 / 16x^2 = -3

[8x^2 (x^2-28) + x^4 - 56x^2 + 784]/ 16x^2 = -3
[8x^2 (x^2-28) + x^4 - 56x^2 + 784] = - 48x^2
8x^4 - 224x^2 +x^4 -56x^2 +784 +48x^2 =0
9x^4 -232x^2 +784 =0
(9x^2-196)(x^2-4)=0
x^2= 196/9 , 4
x=？

y= [(x^2-28)/4x]
自己sub - -

hongji

明天考mathS paper2噢gambate

Allmaths

x^2-4xy=28   and   2xy+y^2=-3

how to find x and y?
-4xy=28-x^2
x=4x^2-7/y用这个方法来做出现x^2  ...
hongji 发表于 13-12-2010 10:31 PM

   
来套比较帅的方法。。。

x^2-4xy=28             --->eq1
y^2+2xy=-3            --->eq 2

eq1+eq2,
x^2-2xy+y^2=25
(x-y)^2=25
x-y=±5
x=y+5   or   y-5

Sub x=y+5 into eq 1,
(y+5)^2-4(y+5)y=28
3y^2+10y+3=0
y=-3 or -1/3
x=2  or 14/3

Sub x=y-5 into eq 1,
(y-5)^2-4(y-5)y=28
3y^2-10y+3=0
y= 3 or 1/3
x=-2 or -14/3


干净又容易。。。

注：有4个solutions

Allmaths

回复  hongji
x^2-4xy=28   ...(1)
y^2+2xy=-3  ...(2)

(2)/(1),
y^2+2xy=-3
_______=_
x^2-4xy  28

rearrange得
28y^2 +56xy=12xy-3x^2
3x^2 +44xy+28y^2=0
(3x+2y)(x+14y)=0
y= -2x/3  or x =-14y

Substitute y=-2x/3 into (1),
...
...
Substitute x=-14y into (1),
...
...
whyyie 发表于 13-12-2010 11:11 PM

   
红字有误。。。

应该是y=-3x/2才对。。。

Allmaths

回复  hongji


[8x^2 (x^2-28) + x^4 - 56x^2 + 784]/ 16x^2 = -3
[8x^2 (x^2-28) + x^4 - 56x^2 + 784] = - 48x^2
8x^4 - 224x^2 +x^4 -56x^2 +784 +48x^2 =0
9x^4 -232x^2 +784 =0
(9x-196)(x-4)=0
x= 196/9 , 4
peaceboy 发表于 13-12-2010 11:14 PM

哥哥，那是x^4，不是quadratic....

hongji

f(x)=x^3 + px^2 +7x + q where p,q are constants.
when x=-1,f'(x)=0 when f(x) is divided by (x+1) the remainder is -16.
Find the value of p & q.
1) Show that f(x) =0 has only one real root.
   Find the set of x such that f(x) >0


2)express [x+q/f(x)] in partial fraction.

我不明白。。><

whyyie

回复 2449# hongji

f(x)=x^3 + px^2 +7x + q
Question 1
f'(x)=3x^2 + 2px +7
f'(-1)=0
f(-1)= -16

=> p = 5, q=-13

找p和q后放进equation
x^3 + 5x^2 + 7x -13=0
(x-1)(x^2 + 6x +13)=0
Since x^2+ 6x + 13 gives complex root.
x=1 is the only real root

f(x)>0
(x-1)(x^2 + 6x +13)>0
x>1

Question 2
[x+q/f(x)] = 1 -13 /[(x-1)(x^2 + 6x +13)]

Let 13 /[(x-1)(x^2 + 6x +13)]be
A/(x-1) + (Bx+c)/(x^2 + 6x +13)
然后solve

hongji

回复  hongji

f(x)=x^3 + px^2 +7x + q
Question 1
f'(x)=3x^2 + 2px +7
f'(-1)=0
f(-1)= -16

...
whyyie 发表于 15-12-2010 09:27 PM

    f'(x)这个我还没学到
所以可以告诉我f'(x)=3x^2+2px+7怎样来的吗？

junchung2003

回复 2451# hongji


    differentiation
初五add math有

hongji

回复  hongji


    differentiation
初五add math有
junchung2003 发表于 15-12-2010 10:26 PM

    我f4有拿addmath但在f5后面临转校。。没学到
    几个科目跟着不一样
    所以没能力和在拿addmath
    我去看chap5吧

junchung2003

回复 2453# hongji


    务必恶补differentiation，没有这个basic，不用考数学

hongji

回复  hongji


    务必恶补differentiation，没有这个basic，不用考数学
junchung2003 发表于 15-12-2010 10:55 PM

我明白。。我已经在看f5的
我的basic差

Lov瑜瑜4ever

An easier way of proving (z1/z2)*=z1*/z2* is to start with (z1/z2)*x(z2)*.
Show how this can be done, remembering that you have already proved that " the conjugate of a product is the product of the conjugates" .

hongji

if z=[(3+2i)/(a+3i)]+[(1+i)/(3+ai)] and Im(z)=R(z) , find the value of a,where a E R

HanQing

An easier way of proving (z1/z2)*=z1*/z2* is to start with (z1/z2)*x(z2)*.
Show how this can be don ...
Lov瑜瑜4ever 发表于 17-12-2010 10:27 AM

尝试~
To prove (z1/z2)*=z1*/z2*,
Consider
(z1/z2)*x(z2)*
={(z1/z2)x(z2)}*the conjugate of a product is the product of the conjugates
=(z1)* &#8596; (z1/z2)*=(z1*/z2*)
看起来好像怪怪的-。-

HanQing

if z=[(3+2i)/(a+3i)]+[(1+i)/(3+ai)] and Im(z)=R(z) , find the value of a,where a  ∈ R

Given z = [(3+2i)/(a+3i)]+[(1+i)/(3+ai)],
Simplifying ,
z= [(3+2i)(3+ai)+(1+i)(a+3i)]/(a+3i)(3+ai)
=(9+6i+3ai-2a+a+3i+ai-3)/(3a+a^2i+9i-3a)
=[(6-a)+(9+4a)i]/(a^2+9)i
=[(9+4a)-(6-a)i]/(a^2+9)
From Im(z)=R(z),
=>(9+4a)/(a^2+9)=-(6-a)(a^2+9)
=>(9+4a)+(6-a)=0
∴a=-5
自己查看，我可能算错。

hongji

if z=[(3+2i)/(a+3i)]+[(1+i)/(3+ai)] and Im(z)=R(z) , find the value of a,where a  ∈ R

Given z =  ...
HanQing 发表于 17-12-2010 04:08 PM

    答案对了