Please write an algebraic expression containing X. no matter what rational number x takes, the value of the algebraic expression is always non negative

Please write an algebraic expression containing X. no matter what rational number x takes, the value of the algebraic expression is always non negative

The column algebra formula is: (x2 + 1) and so on

1) Write an algebraic expression containing the letter X so that no matter what the value of the letter X is, the value of the algebraic expression is always negative
(1) Please write a polynomial containing the letter X (at least 3 terms), so that no matter what the value of the letter X is, the sum of this polynomial is not less than 2011

1) Write an algebraic expression containing the letter X so that no matter what the value of the letter X is, the value of the algebraic expression is always negative
-2x²-4x-3
Please write a polynomial containing the letter X (at least 3 terms), so that no matter what the value of the letter X is, the sum of this polynomial is not less than 2011:1
x²+2x+2012

To write an algebraic expression containing only the letter X, the following requirements are required: (1) to make the algebraic expression meaningful, the letter X must take all the real numbers greater than 1; (2) the value of the algebraic expression is always negative

According to the condition that the quadratic radical is meaningful and the fraction is meaningful, the letter X of 1x − 1 takes all the real numbers greater than 1. If the value of the algebraic formula is always negative, then - 1x − 1 is OK. So the answer is: - 1x − 1

The image of inverse scale function y = K / X passes through point P (m, n), where m, n are the two roots of the equation x ^ 2 + XK + 4 = 0 about X, and the coordinates of point P are obtained

m. N is the two roots of the equation x ^ 2 + XK + 4 = 0 about X,
m+n=-k,mn=4
The image passing through point P (m, n) with inverse scale function y = K / x,
n=k/m
mn=k=4
The solution is m = n = - 2
P(-2,-2)

If there is a point P (2, n) on the image of the inverse scale function y = K / X and its coordinates are two of the quadratic equation T ^ 2-3T + k = 0 with respect to t

2 is a root of the 1-ary quadratic equation T ^ 2-3T + k = 0,
Substituting, the solution is k = 2
Then the solution of the inverse proportional function is y = 2 / X

If f (x) = (m-1) x2 + (m-2) x + (m2-7m + 12) is even function, then the value of M is ()
A. 1B. 2C. 3D. 4

∵ function f (x) = (m-1) x2 + (m-2) x + (m2-7m + 12) is even function, ∵ f (- x) = f (x), ∵ m-1) X2 - (m-2) x + (m2-7m + 12) = (m-1) x2 + (m-2) x + (m2-7m + 12), ∵ m-2 = 0, M = 2, so B

Given the function f (x) = | x-m |, the function g (x) = XF (x) + M & # 178; - 7m. If M = 1, find the solution set of inequality g (x) ≥ 0
Given the function f (x) = | x-m |, the function g (x) = XF (x) + M & # 178; - 7m
(1) If M = 1, find the solution set of inequality g (x) ≥ 0
(2) Finding the minimum value of function g (x) on [3, + ∞)
(3) If for any x1 ∈ (- ∞, 4], there exists x2 ∈ [3, + ∞), such that f (x1) > G (x2) holds, the value range of real number m is obtained

If f (m) = 0, G (x) becomes a constant function
When m = 1, G (x) = - 6, so the solution set of inequality is empty
2. G (x) min = M & # 178; - 7m, because on [3, + ∞), f (x) = | x-m | must be greater than or equal to 0,
So the product of XF (x) is greater than or equal to 0, M & # 178; - 7m is not a variable, but a constant parameter, so g (x) min = M & # 178; - 7m
3. F (x1) > M & # 178; - 7m, that is, | x1-m | M & # 178; - 7m
So x1-m > M & # 178; - 7m or X1 mm & # 178; - 6m or x1

The square of X - the square of Y + MX + 5y-6 can be decomposed into the product of two first-order factors

x^2-y^2+mx+5y-6
=(x^2+mx)-(y^2-5y)-6
=(x+m/2)^2-(y-5/2)^2-m/4+25/4-6
When - M / 4 + 25 / 4-6 = 0, i.e. M = 1
This formula can be divided into
(x + m / 2 + Y-5 / 2) (x + m / 2-y + 5 / 2) (square difference formula)
（x+y-2)(x-y+3)

If the square + X + m of quadratic trinomial x can be decomposed into (x + 4) (x + n), then M = n=

(x+4)(x+n)=x²+(4+n)x+4n=x²+x+m
∴4+n=1 n=-3
m=4n m=-12

If x ^ 2-y ^ 2 + MX + 5y-6 can be decomposed into the product of two first-order factors, find the value of M

M is not equal to 0! It can be written as X * (x + m) - (Y-3) * (Y-2)