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:
1)(2x-3y/7x)-(3x+2y/7y)
2)[1/a(a+1)]+[1/(a+1)(a+2)]+[2/(a+2)(a+3)-[3/a(a+3)]
3) (6x^2/3-x)+(18x/x-3)
F.2 mathz (simplify)
2007-10-01 5:21 am
回答 (2)
2007-10-01 5:51 am
✔ 最佳答案
1)(2x-3y/7x)-(3x+2y/7y)=[(2x-3y)y-(3x+2y)x]/7xy
=[2xy-3y^2-3x^2-2xy]/7xy
= -3(y^2+x^2)/7xy
2)[1/a(a+1)]+[1/(a+1)(a+2)]+[2/(a+2)(a+3)-[3/a(a+3)]
=[(a+2)(a+3)+a(a+3)+2a(a+1)-3(a+1)(a+2)]/[a(a+1)(a+2)(a+3)]
=[(a^2+5a+6)+(a^2+3a)+(2a^2+2a)-(3a^2+9a+6)]/[a(a+1)(a+2)(a+3)]
=[a^2+a]/[a(a+1)(a+2)(a+3)]
=1/[(a+2)(a+3)]
3) (6x^2/3-x)+(18x/x-3)
= [-(6x^2)+(18x)]/x-3
= [-6x(x-3)]/(x-3)
= -6x
2007-10-01 7:12 am
1)(2x-3y/7x)-(3x+2y/7y)
=[(2x-3y)y-(3x+2y)x]/7xy
=[2xy-3y^2-3x^2-2xy]/7xy
= -3(y^2+x^2)/7xy
2)[1/a(a+1)]+[1/(a+1)(a+2)]+[2/(a+2)(a+3)-[3/a(a+3)]
=[(a+2)(a+3)+a(a+3)+2a(a+1)-3(a+1)(a+2)]/[a(a+1)(a+2)(a+3)]
=[(a^2+5a+6)+(a^2+3a)+(2a^2+2a)-(3a^2+9a+6)]/[a(a+1)(a+2)(a+3)]
=[a^2+a]/[a(a+1)(a+2)(a+3)]
=1/[(a+2)(a+3)]
3) (6x^2/3-x)+(18x/x-3)
= [-(6x^2)+(18x)](x-3)
= [-6x(x-3)](x-3)
2007-09-30 23:13:59 補充：
SORRY! I have some mistake above.3) (6x^2/3-x)+(18x/x-3)= [-(6x^2)+(18x)]/x-3= [-6x(x-3)]/(x-3)= -6x
=[(2x-3y)y-(3x+2y)x]/7xy
=[2xy-3y^2-3x^2-2xy]/7xy
= -3(y^2+x^2)/7xy
2)[1/a(a+1)]+[1/(a+1)(a+2)]+[2/(a+2)(a+3)-[3/a(a+3)]
=[(a+2)(a+3)+a(a+3)+2a(a+1)-3(a+1)(a+2)]/[a(a+1)(a+2)(a+3)]
=[(a^2+5a+6)+(a^2+3a)+(2a^2+2a)-(3a^2+9a+6)]/[a(a+1)(a+2)(a+3)]
=[a^2+a]/[a(a+1)(a+2)(a+3)]
=1/[(a+2)(a+3)]
3) (6x^2/3-x)+(18x/x-3)
= [-(6x^2)+(18x)](x-3)
= [-6x(x-3)](x-3)
2007-09-30 23:13:59 補充：
SORRY! I have some mistake above.3) (6x^2/3-x)+(18x/x-3)= [-(6x^2)+(18x)]/x-3= [-6x(x-3)]/(x-3)= -6x
參考: me
收錄日期: 2021-04-13 13:41:59
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070930000051KK05037