1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.

2个回答

  • 1.[x^2-(y-z)^2]/[(x+y)^2-z^2]

    =(x-y+z)(x+y-z)/(x+y+z)(x+y-z)

    =(x-y+z)/(x+y+z)

    2.2/(x-2)+mx/(x^2-4)=3/(x-2)

    mx/(x^2-4)=1/(x-2)

    mx=(x^2-4)/(x-2)

    mx=x+2

    (m-1)x=2

    m=1时,方程会产生增根

    3.(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2)

    2x+3=A(x-1)*(x+2)+Bx*(x+2)+Cx*(x-1)

    2x+3=A(x^2+x-2)+B(x^2+2x)+C(x^2-x)

    2x+3=(A+B+C)x^2+(A+2B-C)x-2A

    A+B+C=0

    A+2B-C=2

    -2A=3

    A=-3/2,B=5/3,C=-1/6

    4.a^2-ab-2b^2=0

    (a-2b)(a+b)=0

    a=2b或a=-b

    a=2b

    a/b-b/a-(a^2+b^2)/ab

    =2-1/2-(4b^2+b^2)/2b^2

    =3/2-5/2

    =-1

    a=-b

    a/b-b/a-(a^2+b^2)/ab

    =-1+1-2b^2/(-b^2)

    =0+2

    =2