计算分式和分式方程(四题)a^2-b^2/a-b/[2+(a^2+b^2/ab)];x^2+1/x^2-1-x-2/x-

1个回答

  • [(a^2-b^2)/(a-b)]/[2+(a^2+b^2/ab)]

    =[(a+b)(a-b)/(a-b)]/[(a^2+2ab+b^2)/ab)]

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

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

    =ab/(a+b)

    (x^2+1)/(x^2-1)-[(x-2)/(x-1)]/[(x-2)/x]

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

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

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

    =-(x-1)/(x+1)(x-1)

    =-1/(x+1)

    1/x+1/(x+1)=5/(2x+2)

    两边乘2x(x+1)

    2x+2+2x=5x

    4x+2=5x

    x=2

    分式方程要检验

    经检验,x=2是方程的解

    (x-8)/(x-7)-1/(7-x)=8

    (x-8)/(x-7)+1/(x-7)=8

    (x-8+1)/(x-7)=8

    (x-7)/(x-7)=8

    1=8

    等式不成立

    所以方程无解