题目内容
已知实数a满足a2+4a-8=0,求
-
•
的值.
| 1 |
| a+1 |
| a+3 |
| a2-1 |
| a2-2a+1 |
| a2+6a+9 |
考点:分式的化简求值
专题:
分析:先根据分式混合运算的法则把原式进行化简,再根据实数a满足a2+4a-8=0得出a2+4a=8代入进行计算即可.
解答:解:原式=
-
•
=
-
=
=
=
∵实数a满足a2+4a-8=0,
∴a2+4a=8
∴原式=
=
.
| 1 |
| a+1 |
| a+3 |
| (a+1)(a-1) |
| (a-1)2 |
| (a+3)2 |
=
| 1 |
| a+1 |
| a-1 |
| (a+1)(a+3) |
=
| a+3-a+1 |
| (a+1)(a+3) |
=
| 4 |
| (a+1)(a+3) |
=
| 4 |
| a2+4a+3 |
∵实数a满足a2+4a-8=0,
∴a2+4a=8
∴原式=
| 4 |
| 8+3 |
| 4 |
| 11 |
点评:本题考查的是分式的化简求值,熟知分式混合运算的法则是解答此题的关键.
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