1个回答
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f(x)=lg(ax-1)-lg(x-1)在区间[10,正无穷)
f(x)=lg(ax-1)-lg(x-1)
=lg
[
(ax-1)/
(x-1)
]
=lg
[[
(a(x-1)
+(a
-1)
]/(x-1)]
=lg
(
a
+
(a-1)/(x-1)]
令:y
=
lgu,
u
=a
+
(a-1)/
(x-1)
u
=a
+
(a-1)/
(x-1)
在【10,正无穷大)上增函数就可以了。
u
(10)=
a
+(a-1)/9
>0
且
a-1
<0
得:1/10
<
a<
1
f(x)=lg(ax-1)-lg(x-1)
=lg
[
(ax-1)/
(x-1)
]
=lg
[[
(a(x-1)
+(a
-1)
]/(x-1)]
=lg
(
a
+
(a-1)/(x-1)]
令:y
=
lgu,
u
=a
+
(a-1)/
(x-1)
u
=a
+
(a-1)/
(x-1)
在【10,正无穷大)上增函数就可以了。
u
(10)=
a
+(a-1)/9
>0
且
a-1
<0
得:1/10
<
a<
1
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