题目内容
(2009•枣庄一模)定义运算法则如下:a?b=a
+b-
,a*b=lga2-lgb
,M=2
?
,N=
*
.
若f(x)=
则f[f(N-
M)]=
.
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
| 1 |
| 4 |
| 8 |
| 125 |
| 2 |
| 1 |
| 25 |
若f(x)=
|
| 2 |
| 9 |
| 1 |
| 4 |
| 1 |
| 4 |
分析:首先由新定义化简N和M,求出N-
M=1-
×4=
,然后求出f(
)=-2,最后把-2代入函数解析式求值.
| 2 |
| 9 |
| 2 |
| 9 |
| 1 |
| 9 |
| 1 |
| 9 |
解答:解:由题意知N=
*
=lg(
)2-lg(
)
=lg2+lg5=1.
M=2
?
=(
)
+(
)-
=
+
=4.
N-
M=1-
×4=
.
又f(x)=
,
所以f(N-
M)=f(
)=log3
=-2.
f(-2)=2-2=
.
故答案为
.
| 2 |
| 1 |
| 25 |
| 2 |
| 1 |
| 25 |
| 1 |
| 2 |
M=2
| 1 |
| 4 |
| 8 |
| 125 |
| 9 |
| 4 |
| 1 |
| 2 |
| 8 |
| 125 |
| 1 |
| 3 |
| 3 |
| 2 |
| 5 |
| 2 |
N-
| 2 |
| 9 |
| 2 |
| 9 |
| 1 |
| 9 |
又f(x)=
|
所以f(N-
| 2 |
| 9 |
| 1 |
| 9 |
| 1 |
| 9 |
f(-2)=2-2=
| 1 |
| 4 |
故答案为
| 1 |
| 4 |
点评:本题是新定义题,考查了指数式的化简与求值,考查了对数式的运算性质,解答的关键是对新定义的理解与运用,是基础题.
练习册系列答案
相关题目