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  • (1)計(jì)算:a-(a^2)/(a+1)-1 (2)計(jì)算:3x/(x-4y)+(x+y)/(4y-x)-7y/(x-4y)

    (1)計(jì)算:a-(a^2)/(a+1)-1 (2)計(jì)算:3x/(x-4y)+(x+y)/(4y-x)-7y/(x-4y)
    (3)計(jì)算:x(2-1/x)+x/x^2-2x*(x^2-4)
    (4)計(jì)算:(x-3)/(2x-4)÷(5/x-2--2)
    (5)計(jì)算:m/m+3-6/9-m^2÷2/m-3
    (6)計(jì)算:(3a/a-3-a/a+3)*a^2-9/a
    數(shù)學(xué)人氣:776 ℃時(shí)間:2020-04-01 14:51:27
    優(yōu)質(zhì)解答
    (1) a-(a²)/(a+1)-1
    =(a²+a-a²-a-1)/(a+1)
    =-1/(a+1)
    (2) 3x/(x-4y)+(x+y)/(4y-x)-7y/(x-4y)
    = (3x-x-y-7y)/(x-4y)
    =(2x-8y)/(x-4y)
    =2
    (3) x(2-1/x)+x/x²-2x*(x²-4)
    =2x-1+1/x-2x³+8x
    =10x-2x³+1/x-1
    (4) (x-3)/(2x-4)÷(5/x-2)
    =(x-3)/(2x-4) *(x-2)/5
    =(x-3)/10
    (5) m/m+3-6/9-m²÷2/m-3
    =m/(m+3)-6/[(m+3)(3-m)]*(m-3)/2
    =m/(m+3)+3/(m+3)
    =(m+3)/(m+3)
    =1
    (6) (3a/a-3-a/a+3)*a²-9/a
    =[3a(a+3)-a(a-3)]/(a²-9) *(a²-9)/a
    =[3a(a+3)-a(a-3)]/a
    =3a+9-a+3
    =2a+12
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