问题
解答题
计算: (1)(
(2)先化简代数式(
|
答案
(1)原式=
•(-c2 a2b2
)•b3 a3c3
=-a4 b4
.1 ab3c
(2)(
+a a+2
)÷2 a-2 1 a2-4
=(
+a a+2
)•(a+2)(a-2)2 a-2
=a(a-2)+2(a+2)
=a2+4.
当a=1,原式=5.
计算: (1)(
(2)先化简代数式(
|
(1)原式=
•(-c2 a2b2
)•b3 a3c3
=-a4 b4
.1 ab3c
(2)(
+a a+2
)÷2 a-2 1 a2-4
=(
+a a+2
)•(a+2)(a-2)2 a-2
=a(a-2)+2(a+2)
=a2+4.
当a=1,原式=5.