已知b/a+b=a+c-b/b+c-a=a+b+c/2a+b+2c且abc不等于0.求分式a+b/c-b的值 步骤详细点! 答案为5
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 18:13:56
已知b/a+b=a+c-b/b+c-a=a+b+c/2a+b+2c且abc不等于0.求分式a+b/c-b的值步骤详细点!答案为5已知b/a+b=a+c-b/b+c-a=a+b+c/2a+b+2c且ab
已知b/a+b=a+c-b/b+c-a=a+b+c/2a+b+2c且abc不等于0.求分式a+b/c-b的值 步骤详细点! 答案为5
已知b/a+b=a+c-b/b+c-a=a+b+c/2a+b+2c且abc不等于0.求分式a+b/c-b的值 步骤详细点! 答案为5
已知b/a+b=a+c-b/b+c-a=a+b+c/2a+b+2c且abc不等于0.求分式a+b/c-b的值 步骤详细点! 答案为5
由于abc≠0,所以b/(a+b)≠0,故可设:
b/(a+b)=(a+c-b)/(b+c-a)=(a+b+c)/(2a+b+2c)=1/k
∴a+b=bk ⑴
b+c-a=ak+ck-bk ⑵
2a+b+2c=ak+bk+ck ⑶
由⑴得:a=(k-1)·b,由a≠0可知k-1≠0,代入⑵得:
c=-(k-2)(k+1)/(k-1)·b
把a、c的值都代入⑶得:
(k-1)(k-2)·b+(k-1)·b-(k-2)^2(k+1)/(k-1)·b=0
解得:k=5/3
∴a=2/3·b,c=4/3·b
∴(a+b)/(c-b)
=(2b/3+b)/(4b/3-b)
=5