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  • (2a^2+3a+2)/(a+1)-(a^2-a-5)/(a+2)-(3a^2-4a-5)/(a-2)+(2a^2-8a+5)/(a-3)

    (2a^2+3a+2)/(a+1)-(a^2-a-5)/(a+2)-(3a^2-4a-5)/(a-2)+(2a^2-8a+5)/(a-3)
    數(shù)學(xué)人氣:306 ℃時(shí)間:2020-05-06 11:07:22
    優(yōu)質(zhì)解答
    =(2a²+2a+a+1+1)/(a+1)-(a²+2a-3a-6+1)/(a+2)-(3a²-6a+2a-4-1)/(a-2)+(2a²-6a-2a+6-1)/(a-3)
    =2a+1+1/(a+1)-a+3-1/(a+2)-3a-2+1/(a-2)+2a-2-1/(a-3)
    =1/(a+1)-1/(a+2)+1/(a-2)-1/(a-3)
    =1/(a+1)(a+2)-1/(a-2)(a-3)
    =[(a-2)(a-3)-(a+1)(a+2)]/ (a+1)(a+2)(a-2)(a-3)
    =(a²-5a+6-a²-3a-2)/(a+1)(a+2)(a-2)(a-3)
    =(4-8a)/(a+1)(a+2)(a-2)(a-3)
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