问题
填空题
已知函数f(x)=
|
答案
∵f(1)=3
∴
=3a+ 1 a 2
∴a+
=6⇒a 1 a
+(1 2
)1 a
=21 2 2
∴f(
)=3 2
=a
+3 2 1 a 3 2 2
(a 1 2
+ (1 2
) 1 a
)(a+1 2
-1)1 a
=
×21 2
×52
=52
故答案为:5
.2
已知函数f(x)=
|
∵f(1)=3
∴
=3a+ 1 a 2
∴a+
=6⇒a 1 a
+(1 2
)1 a
=21 2 2
∴f(
)=3 2
=a
+3 2 1 a 3 2 2
(a 1 2
+ (1 2
) 1 a
)(a+1 2
-1)1 a
=
×21 2
×52
=52
故答案为:5
.2