We know the proposition p: "for any x ∈ R, there exists m ∈ R, 4 ^ x + 2 ^ XM + 1 = 0". If the proposition is not p and is a false proposition, then the value of M of real number Δ > = 0, the solution is m = 2, why m > = 2 is not good

The original title is equivalent to "P is a true proposition" and the range of M is obtained
Let t = 2 ^ x > 0, that is to say, t is a positive number. Have you noticed?
It is not enough to pay attention to the discriminant ⊿ ≥ 0, but it is necessary and only need to omit the negative root (because t > 0, i.e. 2 ^ x > 0)
That is to say, remove the case of M ＜√ (M & # 178; - 4), that is, remove the case of M ≥ 2
The answer should be: m ≤ - 2
I believe you can understand

It is known that P: there exists x ∈ r such that MX2 + 1 ≤ 0; Q: for any x ∈ R, there always exists x2 + MX + 1 > 0. If P or q are false propositions, then the value range of real number m is ()
A. M ≥ 2B. M ≤ - 2C. M ≤ - 2, or m ≥ 2D. - 2 ≤ m ≤ 2

If P is true, then m ＜ 0; if q is true, that is, X2 + MX + 1 ＞ 0, then △ = M2-4 ＜ 0, the solution is - 2 ＜ m ＜ 2. Because P or q are false propositions, then p and Q are all false. So m ≥ 0m ≤ − 2 or m ≥ 2, so m ≥ 2. So a is chosen

If inequality (T ^ 2-2t-3) x - (T-3) X-1

First, make it clear

The equation x × x-mx + 1 is a, B, and a > 0,1

1: According to Weida's theorem
a+b=m
ab=1
m=a+b=b+1/b
2: According to a ^ 2 + B ^ 2 ≥ 2Ab
m=b+1/b≥2
But because B ≠ 1, M > 2
According to the value range of B, we get 2

We know the proposition p: "for any x ∈ R, there exists m ∈ R, 4 ^ x + 2 ^ XM + 1 = 0". If the proposition is not p and is a false proposition, then the value of M of real number Δ > = 0, the solution is m = 2, why m > = 2 is not good