作业帮分式计算 要过程 明天要交大侠救我
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/14 15:34:41
分式计算 要过程 明天要交大侠救我
[(a^2+3)/(a^2-1)]-[(a+1)/(a-1)]+1
要过程,才加10分哦
[(a^2+3)/(a^2-1)]-[(a+1)/(a-1)]+1
要过程,才加10分哦
解:由题可得:
[(a^2+3)/(a^2-1)]-[(a+1)/(a-1)]+1
=(a^2+3)/(a+1)(a-1) - (a+1)^2/(a+1)(a-1) +(a+1)(a-1)/(a+1)(a-1)
=[a^2+3-(a^2+2a+1)+(a+1)(a-1)]/(a+1)(a-1)
=(a^2+3-a^2-2a-1+a^2-1)/(a+1)(a-1)
=(a^2-2a+1)/(a+1)(a-1)
=(a-1)(a-1)/(a+1)(a-1)
=(a-1)/(a+1)
[(a^2+3)/(a^2-1)]-[(a+1)/(a-1)]+1
=(a^2+3)/(a+1)(a-1) - (a+1)^2/(a+1)(a-1) +(a+1)(a-1)/(a+1)(a-1)
=[a^2+3-(a^2+2a+1)+(a+1)(a-1)]/(a+1)(a-1)
=(a^2+3-a^2-2a-1+a^2-1)/(a+1)(a-1)
=(a^2-2a+1)/(a+1)(a-1)
=(a-1)(a-1)/(a+1)(a-1)
=(a-1)/(a+1)