1. The inequality X & # 178; + 2x + 3 ≥ x + m holds for X ∈ [- 2,2], then the value range of M 2. If the inequality | x | + | 2x-3 | - a ＞ 0 holds, the value range of A Let y be a quadratic function, f (x) satisfy f (x + 1) = x & # 178; + X + 1 when x ∈ [- 1,2]. If the inequality f (x) > 2x + m holds, then the value range of real number m is constant

1. The inequality X & # 178; + 2x + 3 ≥ x + m holds for X ∈ [- 2,2], then the value range of M 2. If the inequality | x | + | 2x-3 | - a ＞ 0 holds, the value range of A Let y be a quadratic function, f (x) satisfy f (x + 1) = x & # 178; + X + 1 when x ∈ [- 1,2]. If the inequality f (x) > 2x + m holds, then the value range of real number m is constant

1. The inequality is: x ^ 2 + 2x + 3-x > = m, that is, x ^ 2 + X + 3 > = M
The original problem is to find the minimum value of binomial x ^ 2 + X + 3 in the interval [- 2,2]
The opening of binomial image is upward, and the symmetry axis is - 1 / 2
Because the axis of symmetry is located in the [- 2,2] interval, the minimum value is the vertex, and the ordinate of the vertex is 11 / 4
Therefore, the value range of M is: Ma
That is to find the minimum value of the function | x | + | 2x-3 |
Function into a piecewise function, two key points, one is 0, one is 3 / 2
When x0 and X3 / 2, x + 2x + 3 = 3x-3, the increasing function x = 3 / 2 has a minimum
So, the minimum value of the whole function is 3 / 2,
The value range of a is am
x^2-3x+1>m
This becomes the same type as the first question
Find the minimum value in the interval [- 1,2]
Axis of symmetry: x = 3 / 2, just in the interval, the small value is the vertex
It is - 5 / 4
So a

The solution to the inequality X & # 178; + (1 + m) x + m ≥ 0 (m ∈ R)

x²+(1+m)x+m≥0
(x+1)(x+m)≥0
m∈R
discuss
When M1
X ≤ - m or X ≥ - 1
To sum up
M1, solution set {x | x ≤ - m or X ≥ - 1}
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Solving inequality (m-2) & #178; & gt; 0,

(m-2)²>0
because
(m-2) &# 178; ≥ 0 and take the equal sign when x = 2
therefore
The solution set of inequality is m ≠ 2