计算下列各题:(1)[1/x−3+69−x2−x−16−2x];(2)(x+2yx+y−xy)÷(xy+2−xx−y)

1个回答

  • 解题思路:(1)先将分式通分,再按同分母的分式相加减的法则进行计算即可;

    (2)先算括号里面的,再进行分式的约分即可.

    (1)[1/x−3+

    6

    9−x2−

    x−1

    6−2x];

    =

    2(x+3)

    2(x2−9)-[12

    2(x2−9)+

    (x−1)(x+3)

    2(x2−9),

    =

    2x+6−12+x2+3x−x−3

    2(x2−9),

    =

    x2+4x−9

    2(x2−9);

    (2)(

    x+2y/x+y−

    x

    y)÷(

    x

    y+2−

    x

    x−y),

    =

    y(x+2y)−x(x+y)

    y(x+y)]÷

    x(x−y)+2y(x−y)−xy

    y(x−y),

    =

    2y2−x2

    y(x+y)•

    y(x−y)

    x2−2y2,

    =-[x−y/x+y].

    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,要注意运算顺序.