问题
解答题
计算: (1)
(2)
|
答案
(1)原式=2a+2 a+1
=2;
(2)原式=
×(y-2)2 2(y-3)
÷1 y+3 6(2-y) (3-y)(3+y)
=
×(y-2)2 2(y-3)(y+3) (3-y)(3+y) 6(2-y)
=
.y-2 12
计算: (1)
(2)
|
(1)原式=2a+2 a+1
=2;
(2)原式=
×(y-2)2 2(y-3)
÷1 y+3 6(2-y) (3-y)(3+y)
=
×(y-2)2 2(y-3)(y+3) (3-y)(3+y) 6(2-y)
=
.y-2 12