Given f (x) = loga (x + 1-A), find the set of values that make f (x) > 1

When 0a-1,
And x > 2a-1 > A-1, so the solution set is x > 2a-1
When a > 1, loga (x + 1-A) is an increasing function,
f(x)

1、 Find the definition fields of the following functions: (1) y = loga 1 / x-3; (2) y = loga (the square of 1-x),

（1）y=loga 1/x-3
1. X is not equal to 3
2.1/（x-3）>0,x-3>0,x>3
The domain of the function is x > 3
(2) Y = loga (square of 1-x)
The square of 1-x is greater than 0
(1+x)(1-x)>0
1. X1 (rounding off)
2.-1

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