化簡多項式,help!!!

2007-11-16 2:05 am
1......( 3 x ) / ( x ^ 2 - 1) 乘 { x + 2 + ( 1 / x ) }

2......x / { y - ( y ^ 2 / x ) } 加 y / { x - (x ^ 2 / y ) }

3......2 / ( a ^ 2 - 1 )減 1 / { ( a + 1 ) ^ 2 } 減 1 / { (a - 1) ^ 2 }

4......2 / ( x ^ 2 - x ) 減 1 / ( x ^ 3 - x )

回答 (3)

2007-11-16 2:35 am
✔ 最佳答案
1) [3x / (x^2 - 1)] * [x + 2 + (1/x)]
= {3x / [(x + 1)*(x - 1)]} * [(x^2 + 2x + 1) / x]
= {3x / [(x + 1)*(x - 1)]} * {[(x + 1)*( x + 1)] / x}
= 3 * (x + 1) / (x - 1)

2) x / {y - [(y^2) / x]} + y / {x - [(x^2) / y]}
= [(x^2) / (xy - y^2)] + [(y^2) / (xy - x^2)]
= {(x^2) / [y*(x - y)]} + {(y^2) / [x*(y - x)]}
= {(x^2) / [y*(x - y)]} - {(y^2) / [x*(x - y)]}
= {(x^3) / [xy*(x - y)]} - {(y^3) / [xy*(x - y)]}
= (x^3 - y^3) / [xy*(x - y)]
= [(x - y) * (x^2 + xy + y^2)] / [xy * (x - y)]
= (x^2 + xy + y^2) / xy

3) [2 / (a^2 - 1)] - {1 / [( a + 1 )^2]} - {1 / [(a - 1)^2]}
= {[2*(a + 1)*(a - 1)] - [(a - 1)^2] - [(a + 1)^2]} / {[(a + 1)^2]*[(a - 1)^2]}
= -1 * [(a + 1) - (a - 1)]^2 / {[(a + 1)^2]*[(a - 1)^2]}
= -1 * (2)^2 / {[(a + 1)^2]*[(a - 1)^2]}
= -4 / {[(a + 1)^2]*[(a - 1)^2]}

4) [2 / (x^2 - x)] - [1 / (x^3 - x)]
= {2 / [x * (x - 1)]} - {1 / [x * (x^2 - 1)]}
= {2 / [x * (x - 1)]} - {1 / [x * (x - 1) * (x + 1)]}
= [2 * (x + 1) - 1] / [x * (x - 1) * (x + 1)]
= (2x - 1) / [x * (x - 1) * (x + 1)]
2007-11-16 3:04 am
1)( 3 x ) / ( x ^ 2 - 1) X { x + 2 + ( 1 / x ) }
=[(3x) / (x^2-1)][(x^2+2x+1)/x]
=[(3x) / (x^2-1)]{[(x+1)^2]/x}
=[3(x+1)]/(x-1)x
=[3(x+1)]/x^2-x

2) x / { y - ( y ^ 2 / x ) } + y / { x - (x ^ 2 / y ) }
=x / [y(1-y/x)] + y / [x(1-x/y)]
=x / y[(x-y)/x] + y / x[(y-x)/y]
=x / (y/x)(x-y) - y / (x/y)(x-y)
=[x(x/y) - y(y/x)] / (y/x)(x/y)(x-y)
=[(x^2)/y - (y^2)/x] / (x-y)






3)2 / ( a ^ 2 - 1 ) - 1 / { ( a + 1 ) ^ 2 } - 1 / { (a - 1) ^ 2 }
=2/[(a+1)(a-1)] - 1/(a+1)^2 - 1/(a-1)^2
=[2(a+1)(a-1)-(a-1)^2-(a+1)^2] / [(a+1)^2][(a-1)^2]
=(2a^2 - 2 - a^2 - 1 + 2a - a^2 - 1 - 2a) / [(a+1)^2(a-1)^2]
= - 4 / (a+1)^2(a-1)^2

4)2 / ( x ^ 2 - x ) - 1 / ( x ^ 3 - x )
=2/[x(x-1)] - 1/[x(x^2-1)]
=2/[x(x-1)] - 1/[x(x+1)(x-1)]
=[2(x+1) - 1]/[x(x+1)(x-1)]
=(2x+1) / (x)(x+1)(x-1)
參考: 打左好耐..應該沒錯的...希望你睇得明啦^^
2007-11-16 2:17 am
1. [ 3x / x^2-1 ] x [ x+2 + (1/x) ]
=[ 3x / (x+1)(x-1) ] x { [x(x+2)]/x + 1/x }
=[ 3x / (x+1)(x-1) ] x [ (x^2x+1)/x ]
=[ 3x / (x+1)(x-1) ] x [ (x+1)(x+1) /x ]
= 3x(x+1) / x(x-1)
= (2x^2 + 2x) / x-1

2007-11-15 18:26:17 補充:
4. 2/(x^2 - x) - [ 1/( x^3 - x) ] =2/ x(x-1) - 1/ x(x^2 -1) =[ 2(x^2 -1) - (x-1) ] / x(x-1)(x^2 -1) =( 2x^2 -2-x+1 ) / x(x-1)(x^2 -1) =(2x+1)(x-1) / x(x-1)(x^2 -1) =2x+1 / x(x^2 -1)


收錄日期: 2021-04-13 22:25:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071115000051KK02368

檢視 Wayback Machine 備份