If f (x) = 9 to the x power and (K + 1) 3 to the x power + 1 > 0 holds, then the range of K is

If f (x) = 9 to the x power and (K + 1) 3 to the x power + 1 > 0 holds, then the range of K is

f(x)=9^x-(k+1)3^x+1=(3^x)^2-(k+1)3^x+1=[3^x-(k+1)/2]^2+1-[(k+1)/2]^2
1. If (K + 1) / 2 > 0, that is k > - 1, then
f(x)=[3^x-(k+1)/2]^2+1-[(k+1)/2]^2≥1-[(k+1)/2]^2
If f (x) > 0 is constant, there must be
1-[(k+1)/2]^2>0
-1

The x power of 2 + the x power of (k-1} / 2 > 0 is constant when [0, positive infinity], and the value range of K is obtained
Detailed process

2 ^ x + (k-1) / 2 ^ x > 0, where x > = 0
That is, (2 ^ x) ^ 2 + k-1 > 0
k> 1 - (2 ^ x) ^ 2 is constant, while 2 ^ x > = 1
So (2 ^ x) > = 1,1 - (2 ^ x) ^ 20

Given that f (x) = 2x power of 3 - x power of K * 3 + 2, when x ∈ R, f (x) is always positive, and the value range of K?

f(x)=3^(2x)-k*3^x+2>0
3^x(3^x+2/3^x-k)>0
Because 3 ^ x > 0
therefore
3^x+2/3^x-k>0
∵3^x+2/3^x≥2√2
And K

Given that x ^ 2 + y ^ 2 = 1, if x + Y-K ≥ 0 holds for any x, then K max?

-√2
It is equivalent to finding the minimum value of X + y
x+y=cosa+sina=√2sin（a+45°）
The minimum is naturally - √ 2