有几道分式的计算题不会做计算题有两个:(1/m²-m)+(m-5/2m²-2)((a/a-b)+(b

1个回答

  • 计算题:

    [1/(m²-m)]+[(m-5)/(2m²-2)]

    ={ 1/[m(m-1)] }+{ (m-5)/[2(m+1)(m-1)] } 通分,最简公分母是2m(m+1)(m-1)

    ={ 2(m+1)/[2m(m+1)(m-1)] }+{ m(m-5)/[2m(m+1)(m-1)] }

    =[2(m+1)+m(m-5)]/[2m(m+1)(m-1)]

    =(2m+2+m²-5m)/[2m(m+1)(m-1)]

    =(m²-3m+2)/[2m(m+1)(m-1)]

    =(m-1)(m-2)/[2m(m+1)(m-1)]

    =(m-2)/[2m(m+1)]

    [a/(a-b)+b/(b-a)]÷[1/(a+b)]

    =[a/(a-b)-b/(a-b)]×(a+b)

    =[(a-b)/(a-b)]×(a+b)

    =1×(a+b)

    =a+b

    解方程:

    [1/(x+2)]+[1/(x-2)]=4/(x²-4) 方程两边同时乘最简公分母(x²-4)

    x-2+x+2=4

    2x=4

    x=2

    检验:把x=2代入x²-4=2²-4=4-4=0

    经检验x=2是原方程的增根,原方程无解

    [3x/(x+2)]+[2/(x-2)]=3 方程两边同时乘最简公分母(x+2)(x-2)

    3x(x-2)+2(x+2)=3(x+2)(x-2)

    3x²-6x+2x+4=3x²-12

    -6x+2x=-12-4

    -4x=-16

    x=4

    检验:把x=4代入(x+2)(x-2)=(4+2)(4-2)=12≠0

    经检验x=4是原方程的根

    (1-x)/(x-2)=[1/(2-x)]-2 方程两边同时乘最简公分母(x-2)

    1-x=-1-2(x-2)

    1-x=-1-2x+4

    -x+2x=-1+4-1

    x=2

    检验:把x=2代入x-2=2-2=0

    经检验x=2是原方程的增根,原方程无解

    3/2-[1/(3x-1)]=5/(6x-2) 方程两边同时乘最简公分母2(3x-1)

    3(3x-1)-2=5

    9x-3-2=5

    9x=5+3+2

    9x=10

    x=10/9

    检验:把x=10/9代入2(3x-1)=2(3×10/9-1)=2×(10/3-1)=14/3≠0

    经检验x=10/9是原方程的根