f.2 introduction to polynomial

2010-12-29 6:08 am
1)a)i)evaluate (-1)^2,(-1)^3,(-1)^4
ii)Hence find the values of (-1)^2n and (-1)^2n+1,where n is a positive integer.
b)Using the results of (a)(ii),evaluate (-1)^1996-(-1)^2057 / (-1)^37*4321
2)(v-2u)^2-(2u-v)(u+2v)-2v+4u
3)2(2-a)(-1+2a)+4(a-2)(2+a)
4)2x(-1+2a)-4x(2+a)
5)If x^4p+1/x^p+2=x^3p+5/y ,where p is a positive integer,express y in terms od x.
更新1:

re 知識就是力量 : i don't know how to explain, but i can show the question to you: http://s002.radikal.ru/i197/1012/7e/e3c0a969f877.jpg

回答 (3)

2010-12-29 7:58 am
✔ 最佳答案
(-1)^2=1
(-1)^3=-1
(-1)^4=1
_________
(-1)^2n=1
(-1)^2n+1=-1
_____________________
(-1)^1996-(-1)^2057 / (-1)^37*4321
(1-(-1))/(-1)
2/-1
-2
_________________________________
(v-2u)^2-(2u-v)(u+2v)-2v+4u
(2u-v)^2-(2u-v)(u+2v)+2(2u-v)
(2u-v)(2u-v-u-2v+2u-v)
(2u-v)(3u-4v)
________________________
2(2-a)(-1+2a)+4(a-2)(2+a)
(2-a)(4a-2)-(2-a)(8+4a)
(2-a)(4a-2-8-4a)
(2-a)(-10)
____________________
2x(-1+2a)-4x(2+a)
2x(2a-1-2(2+a))
2x(2a-1-4-2a)
2x(-5)
-10x
__________________

2010-12-29 00:00:33 補充:
(x^(4p+1))/(x^(p+2))=(x^(3p+5))/y
(x^p+2)(x^(3p+5))=y(x^(4p+1))
x^(4p+7)=y(x^(4p+1))
(x^4p+7)/(x^4p+1)=y
x^6=y
y=x^6
2010-12-29 6:44 am
i don't know how to explain, but i can show to you:
http://s002.radikal.ru/i197/1012/7e/e3c0a969f877.jpg
2010-12-29 6:22 am
I don't understand the question 5..
it is completely a mess.

2010-12-28 22:23:47 補充:
do you mean:
[(x^4)*p + 1] / [(x^p + 2)] = [(x^3)*p + 5] / y


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