The solution of the equation (t-1) / (T-2) = (2a-3) (a + 5) about t is equal to 0 RT

(t-1) / (T-2) = (2a-3) (a + 5) the solution is equal to 0, that is, t = 0
Substituting: - 1 / - 2 = (2a-3) (a + 5)
The results are as follows
4a^2+14a-31=0
The solution is as follows
a=1.54
or
a=-5.04

If the square of negative triple 2a-1 is equal to 0, then a is equal to?, and the solution of the equation is x equal to?

-Is 3 (2a-1) ^ 2 = 0 the formula?
Where is x? Can I send pictures?

If a is less than or equal to B, compare the size of - 2A + 5 with that of 5-2a, and explain the reason

The known function f (x) = [1-4 / (2a ^ x + a)] (a > 0 and a is not equal to 1) is an odd function defined on R
(1) When x belongs to (0,1 >, TF (x) > = 2 ^ X-2 is constant, the value range of real number T is obtained
When I do it, I first discuss the interval in the stem, divide the denominator to the right of the inequality, and then calculate the derivative Max to make t greater than or equal to 3
Because it's an odd function and a symmetric interval, I do the same thing
THANKS

1. Because f (x) is an odd function defined on R
So f (0) = 0
Then a = 2
two
So f (x) = 1 - 2 / (2 ^ x + 1)
Because 2 ^ x > 0, 2 ^ x + 1 > 1,
So 0 < 2 / (2 ^ x + 1) < 2
So 0 ＞ - 2 / (2 ^ x + 1) ＞ - 2
So 1 ＞ 1 - 2 / (2 ^ x + 1) ＞ - 1
So the range is (- 1,1)
f(x) = 1 - 2/(2^x + 1) = (2^x-1)/（2^x+1)
TF (x) ≥ 2 ^ X-2 means t (2 ^ x-1) / (2 ^ x + 1) ≥ 2 ^ X-2
That is t ≥ (2 ^ x + 1) (2 ^ X-2) / (2 ^ x-1)
=[(2^x-1)^2 + (2^x-1) - 2]/(2^x-1)
=(2^x-1) + 1 - 2/(2^x-1)
If you want to be constant, it must be greater than its maximum value
When x belongs to (0,1), (2 ^ x-1) is an increasing function, - 2 / (2 ^ x-1), it is also an increasing function, so (2 ^ x-1) + 1 - 2 / (2 ^ x-1) is an increasing function, so it is the maximum value when x = 1
Here = 2 - 1 + 1 - 2 / (2 - 1) = 0
So just t > 0
So the range of T is (0, + ∞)

The solution of the equation (t-1) / (T-2) = (2a-3) (a + 5) about t is equal to 0 RT