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  • 1.a-2分之a(chǎn)+1·a分之a(chǎn)-1-a^2-2a分之4a-5 2.x^2-x分之x^2+2x+1·x+1分之x-x-1分之1

    1.a-2分之a(chǎn)+1·a分之a(chǎn)-1-a^2-2a分之4a-5 2.x^2-x分之x^2+2x+1·x+1分之x-x-1分之1
    3.a-2分之a(chǎn)+3-a分之1÷9-a^2分之2-a.其中a=2分之5
    數(shù)學(xué)人氣:181 ℃時(shí)間:2020-04-29 04:29:01
    優(yōu)質(zhì)解答
    (a+1)/(a-2)*(a-1)/a-(4a-5)/(a^2-2a)
    =(a+1)(a-1)/[a(a-2)]-(4a-5)/[a(a-2)]
    =(a^2-1)/[a(a-2)]-(4a-5)/[a(a-2)]
    =[(a^2-1)-(4a-5)]/[a(a-2)]
    =(a^2-1-4a+5)/[a(a-2)]
    =(a^2-4a+4)/[a(a-2)]
    =(a-2)^2/[a(a-2)]
    =(a-2)/a
    (x^2+2x+1)/(x^2-x)*x/(x+1)-1/(x-1)
    =(x+1)^2/[x(x-1)]*x/(x+1)-1/(x-1)
    =(x+1)/(x-1)-1/(x-1)
    =(x+1-1)/(x-1)
    =x/(x-1)
    (a+3)/(a-2)-1/a÷(2-a)/(9-a^2)
    =(a+3)/(a-2)-1/a÷(a-2)/(a^2-9)
    =(a+3)/(a-2)-1/a*(a^2-9)/(a-2)
    =a(a+3)/[a(a-2)]-(a^2-9)/[a(a-2)]
    =[a(a+3)-(a^2-9)]/[a(a-2)]
    =[a^2+3a-a^2+9]/[a(a-2)]
    =(3a+9)/[a(a-2)]
    =3(a+3)/[a(a-2)]
    =3*(5/2+3)/[5/2*(5/2-2)]
    =(3*11/2)/(5/2*1/2)
    =(33/2)/(5/4)
    =33/2*4/5
    =66/5hey 還有一道題目, x-1 分之2 + 1- x分之()=1 填空,我多加你10分!2/(x-1)+()/(1-x)=1 2/(x-1)-()/(x-1)=1 [2-()]/(x-1)=1 2-()=x-1 ()=2+1-x ()=3-x
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