早教吧作业答案频道 -->数学-->
化简b-c/(a-b)(a-c)+c-a/(b-c)(b-a)+a-b/(c-a)(c-b)+2/b-a-2/c-a
题目详情
化简b-c/(a-b)(a-c)+c-a/(b-c)(b-a)+a-b/(c-a)(c-b)+2/b-a-2/c-a
▼优质解答
答案和解析
(b-c)/(a-b)(a-c)+(c-a)/(b-c)(b-a)+(a-b)/(c-a)(c-b)+2/(b-a)-2/(c-a)
=(b-c)/(a-b)(a-c)+(a-c)(b-c)/(a-b)+(a-b)/(a-c)(b-c)-2/(a-b)+2/(a-c)
=[(b-c)²+(a-c)²+(a-b)²]/(a-b)(a-c)(b-c)+[2(a-b)-2(a-c)]/(a-c)(a-b)
=[(b-c)²+(a-c)²+(a-b)²]/(a-b)(a-c)(b-c)-2(b-c)/(a-c)(a-b)
=[(b-c)²+(a-c)²+(a-b)²-2(b-c)²]/(a-b)(a-c)(b-c)
=[(a-c)²+(a-b)²-(b-c)²]/(a-b)(a-c)(b-c)
=[(a-c)²+(a-b+b-c)(a-b-b+c)]/(a-b)(a-c)(b-c)
=(a-c)(a-c+a+c-2b)/(a-b)(a-c)(b-c)
=2(a-b)/(a-b)(b-c)
=2/(b-c)
=(b-c)/(a-b)(a-c)+(a-c)(b-c)/(a-b)+(a-b)/(a-c)(b-c)-2/(a-b)+2/(a-c)
=[(b-c)²+(a-c)²+(a-b)²]/(a-b)(a-c)(b-c)+[2(a-b)-2(a-c)]/(a-c)(a-b)
=[(b-c)²+(a-c)²+(a-b)²]/(a-b)(a-c)(b-c)-2(b-c)/(a-c)(a-b)
=[(b-c)²+(a-c)²+(a-b)²-2(b-c)²]/(a-b)(a-c)(b-c)
=[(a-c)²+(a-b)²-(b-c)²]/(a-b)(a-c)(b-c)
=[(a-c)²+(a-b+b-c)(a-b-b+c)]/(a-b)(a-c)(b-c)
=(a-c)(a-c+a+c-2b)/(a-b)(a-c)(b-c)
=2(a-b)/(a-b)(b-c)
=2/(b-c)
看了 化简b-c/(a-b)(a-...的网友还看了以下:
设a=(√5-1)/2,求(a^5+a^4-2a^3-a^2-a+2)/a^3-a∵2a=√5-1 2020-04-05 …
若a+b=b+c,则a-b(c为整式)若a=b,则ac=bc(c为整式)若ac=bc,则a=b(c 2020-04-22 …
两道集合论的题1设ABC是全集U的任意子集.a)若A∩B=A∩C,A∩B=~A∩C,证明:B=Cb 2020-06-07 …
在三角形ABC中已知(a+b)/a=...在三角形ABC中已知(a+b)/a=sinB/(sinB 2020-06-12 …
已知抛物线y1=ax2+bx+c(a≠0,a≠c)过点A(1,0),顶点为B,且抛物线不经过第三象 2020-07-26 …
一、已知数集M满足条件:若a∈M,则(1+a)/(1-a)∈M(a≠0,a≠±1)(1)若3∈M, 2020-07-30 …
直角三角形ABC中,BC=2,AC=6,依下列的步骤抄作折纸.(A)将A,C两点重合(B)DE为折痕 2020-11-06 …
已知三角形ABC中角ABC所对边分别为abc若ABC成等差数列b=2记角A=x,a+c=fx1已知三 2020-11-24 …
直角三角形ABC中,BC=2,AC=6,依下列的步骤抄作折纸.(A)将A,C两点重合(B)DE为折痕 2020-12-02 …
分类讨论是一种重要的数学方法,如在化简|a|时,可以这样分类:当a>0时,|a|=a;当a=0时,| 2020-12-27 …