先化简再求值:(ab+a)/(b²-1)+(b-1)/(b²-2b+1),其中根号(b-2)+36a²+b²-12ab=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 03:15:52
![先化简再求值:(ab+a)/(b²-1)+(b-1)/(b²-2b+1),其中根号(b-2)+36a²+b²-12ab=0](/uploads/image/z/432228-12-8.jpg?t=%E5%85%88%E5%8C%96%E7%AE%80%E5%86%8D%E6%B1%82%E5%80%BC%EF%BC%9A%28ab%2Ba%29%2F%EF%BC%88b%26%23178%3B-1%EF%BC%89%2B%EF%BC%88b-1%29%2F%28b%26%23178%3B-2b%2B1%EF%BC%89%2C%E5%85%B6%E4%B8%AD%E6%A0%B9%E5%8F%B7%EF%BC%88b-2%29%2B36a%26%23178%3B%2Bb%26%23178%3B-12ab%3D0)
先化简再求值:(ab+a)/(b²-1)+(b-1)/(b²-2b+1),其中根号(b-2)+36a²+b²-12ab=0
先化简再求值:(ab+a)/(b²-1)+(b-1)/(b²-2b+1),其中根号(b-2)+36a²+b²-12ab=0
先化简再求值:(ab+a)/(b²-1)+(b-1)/(b²-2b+1),其中根号(b-2)+36a²+b²-12ab=0
|b-2|+36a²+b²-12ab=0可化为:
|b-2|+(6a-b)²=0
则有:b-2=0且6a-b=0
解得:b=2,a=b/6=1/3
所以:
(ab+a)/(b²-1)+(b-1)/(b²-2b+1)
=a(b+1)/[(b+1)(b-1)] + (b-1)/(b-1)²
=a/(b-1) + 1/(b-1)
=(a+1)/(b-1)
=(1/3 +1)/(2-1)
=4/3