分式计算 要过程 明天要交大侠救我
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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
![分式计算 要过程 明天要交大侠救我](/uploads/image/z/19223442-18-2.jpg?t=%E5%88%86%E5%BC%8F%E8%AE%A1%E7%AE%97+%E8%A6%81%E8%BF%87%E7%A8%8B+%E6%98%8E%E5%A4%A9%E8%A6%81%E4%BA%A4%E5%A4%A7%E4%BE%A0%E6%95%91%E6%88%91)
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)