[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 18:42:16
![[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简](/uploads/image/z/12055918-22-8.jpg?t=%5B%28s%2Fa%2Bs%2Fb%29%2F2%5D%2F%5B%28s%2Fa-s%2Fb%29%2F2%5D%E6%80%8E%E4%B9%88%E5%8C%96%E7%AE%80%5B%EF%BC%88s%2Fa%2Bs%2Fb%EF%BC%89%2F2%5D%2F%5B%EF%BC%88s%2Fa-s%2Fb%EF%BC%89%2F2%5D%E6%80%8E%E4%B9%88%E5%8C%96%E7%AE%80)
[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简
[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简
[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简
[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简[(s/a+s/b)/2]/[(s/a-s/b)/2]怎么化简
原式=(s/a+s/b)/(s/a-s/b)=(a+b)s/(b-a)s=(a+b)/(b-a)
先同分