F.2 mathz help plz~

2007-09-23 8:18 am
plz help me to answer the question,if you do not know the answer,plz answer no~i need the step ~thank for your help~

simplify the following:
1) (1/x-1)-(1/x)

2) (x/x+y)+(y/x-y)

3) [x+1/x(x+2)]-(1/x+2)

4) [2x/(x+1)(x-2)]+(3/2-x)

回答 (2)

2007-09-23 8:49 am
✔ 最佳答案
1) [ 1 / ( x - 1 ) ] - ( 1 / x )

= [ x / x ( x - 1 ) ] - ( x - 1 ) / x ( x - 1 ) [ The L.C.M. of x and ( x - 1 ) is x ( x - 1 )]

= ( x - x + 1 ) / x ( x - 1 )

= 1 / x ( x - 1 )

2) [ x / ( x + y ) ] + [ y / ( x - y )]

= [x ( x - y ) / ( x + y )( x - y ) ] + [ y ( x + y ) / ( x - y )( x + y ) ]

[ The L.C.M. of ( x + y ) and ( x - y ) is ( x - y )( x + y )]

= [ x ( x - y ) + y ( x + y ) ] / ( x + y )( x - y )

= ( x^2 - xy + xy + y^2 ) / ( x + y )( x - y )

= ( x^2 + y^2 ) / ( x + y )( x - y )

3) [ ( x + 1 ) / x ( x + 2 ) ] - 1 / ( x + 2 )

= [ ( x + 1 ) / x ( x + 2 ) ] - x / x ( x + 2 ) [ The L.C.M. of x and ( x + 2 ) is x ( x + 2 ) ]

= [ ( x + 1 ) - x ] / x ( x + 2 )

= 1 / x ( x + 2 )

4) [2x / ( x + 1 )( x - 2 ) ] + 3 / ( 2 - x )

= [ 2x / ( x + 1 )( x - 2 ) ] - 3 / ( x - 2 ) [ Take out the -VE sign in ( 2 - x )]

= [ 2x / ( x + 1 )( x - 2 ) ] - 3 ( x + 1 ) / ( x - 2 )( x + 1 )

= [ 2x - 3 ( x + 1 ) ] / ( x - 2 )( x + 1 )

[ The L.C.M. of ( x - 2 ) and ( x - 1 ) is ( x - 2 )( x - 1 )]

= ( 2x - 3x - 3 ) / ( x - 2 )( x + 1 )

= ( - x - 3 ) / ( x - 2 )( x + 1 )

= - ( x + 3 ) / ( x - 2 )( x + 1 )
參考: My Maths Knowledge
2007-09-23 8:31 am
無得計

因為你寫得唔清楚


收錄日期: 2021-04-13 13:35:13
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070923000051KK00092

檢視 Wayback Machine 備份