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  • 若m^2+3m-1=0,則【(m-3)/(3m^2-6)】除以【(m+2)-(5)/(m-2)】

    若m^2+3m-1=0,則【(m-3)/(3m^2-6)】除以【(m+2)-(5)/(m-2)】
    數(shù)學人氣:754 ℃時間:2020-01-28 06:25:04
    優(yōu)質解答
    m^2+3m-1=0,m^2=1-3m
    【(m-3)/(3m^2-6)】除以【(m+2)-(5)/(m-2)】
    =[(m-3)/(3(m^2-2))] / [(m+2)-5/(m-2)]
    =[(m-3)/(3(1-3m-2))] / [(m^2-4-5)/(m-2)]
    =-[(m-3)/(3(3m+1))] / (m+3)(m-3)/(m-2)]
    =-[(m-3)/(3(3m+1))] * [(m-2)/((m+3)(m-3))]
    =-(m-2)/[3(3m+1)(m+3)]
    =-(m-2)/[3(3m^2+10m+3)]
    =-(m-2)/[3(3-9m+10m+3)]
    =-(m-2)/[3(m+6)]..親,實在抱歉,應該是若m^2+3m-1=0,則【(m-3)/(3m^2-6m)】除以【(m+2)-(5)/(m-2)】你這不是害人嘛m^2+3m-1=0, m^2+3m=1, m(m+3)=1【(m-3)/(3m^2-6m)】除以【(m+2)-(5)/(m-2)】=[(m-3)/(3m(m-2))] / [(m+2)-5/(m-2)]=[(m-3)/(3m(m-2))] / [(m^2-4-5)/(m-2)]=[(m-3)/(3m(m-2))] / (m+3)(m-3)/(m-2)]=[(m-3)/(3m(m-2))] * [(m-2)/((m+3)(m-3))]=1/[3m(m+3)]=1/3
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