问题 解答题
(1)计算
1
2
+1
-
8
+(
3
-1)0

(2)已知实数a满足a2+2a-8=0,求
1
a+1
-
a+3
a2-1
×
a2-2a+1
a2+4a+3
的值.
答案

(1)

1
2
+1
-
8
+(
3
-1)0=
2
-1-2
2
+1

=-

2

(2)

1
a+1
-
a+3
a2-1
×
a2-2a+1
a2+4a+3

=

1
a+1
-
a+3
(a-1)(a+1)
×
(a-1)2
(a+1)(a+3)

=

1
a+1
-
a-1
(a+1)2
=
2
(a+1)2

由已知,实数a满足a2+2a-8=0,故(a+1)2=9,

∴原式=

2
9
(9分).

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