计算:(1)[2aa2−4+1/2−a](2)(−ab)2÷(2a25b)2•a5b(3)a2a−1−a−1(4)(xx

1个回答

  • 解题思路:(1)先化成同分母的分式,再相加即可;

    (2)按运算顺序,先算乘方,再算乘除,最后约分即可;

    (3)将整式的分母看作1,化成同分母的分式,再相加即可;

    (4)按运算顺序,先乘除,再算加减,有括号的,先算括号里面的,最后约分即可.

    (1)[2a

    a2−4+

    1/2−a]=[2a

    (a+2)(a−2)-

    a+2

    (a+2)(a−2),

    =

    2a−a−2

    (a+2)(a−2),

    =

    1/a+2];

    (2)(

    −a

    b)2÷(

    2a2

    5b)2•

    a

    5b,

    =

    a2

    b2•

    25b2

    4a4•[a/5b],

    =[5/4ab];

    (3)

    a2

    a−1−a−1=

    a2

    a−1-(a+1),

    =

    a2

    a−1-

    a2−1

    a−1,

    =

    a2−a2+1

    a−1,

    =[1/a−1];

    (4)(

    x

    x−y−

    2y

    x−y)•

    xy

    x−2y÷(

    1

    x+

    1

    y),

    =[x−2y/x−y]•[xy/x−2y]•[xy/x+y],

    =

    x2y2

    x2−y2.

    点评:

    本题考点: 分式的混合运算.

    考点点评: 本题是计算题,比较简单,考查了分式的混合运算,一定要注意运算的顺序.