It is known that the inequality 2x + 2x − a ≥ 7 on X ∈ (a, + ∞) holds, then the minimum value of real number a is () A. 32B. 1C. 2D. 52

It is known that the inequality 2x + 2x − a ≥ 7 on X ∈ (a, + ∞) holds, then the minimum value of real number a is () A. 32B. 1C. 2D. 52

∵ the inequality 2x + 2x − a ≥ 7 on X ∈ (a, + ∞) is tenable, ∵ x ＞ a, ∵ y = 2x + 2x − a = 2 (x-a) + 2x − a + 2A ≥ 22 (x − a) · 2x − a + 2A = 4 + 2a, if and only if 2 (x − a) = 2x − a, that is, x = a + 1, take the equal sign, ∵ 2x + 2x − a) min = 4 + 2
Known polynomials x ` 2009-x ` 2008y + X ` 2007y ` 2-x ` 2006y ` 3 + ··· XY ` 2008-y ` 2009
(1) Say the number of terms and degree of this polynomial
(2) Write items 100 and 101 from front to back
(3) If x = 1, y = - 1, find the value of this polynomial
1. Number of items = 2010, times = 2009
2. Item 100: - x ^ 1910y ^ 99; item 101: x ^ 1909y ^ 100, just write according to the number of Y
3、x=1,y=-1
Original formula = 1 + 1 + 1 + +1=2010
It is known that the univariate quadratic equation x & sup2; + (2m-1) x + M & sup2; = 0 has two real roots X1 and x2
(1) The range of the real root of M is obtained;
(2) When x 1 & sup 2; - x 2 & sup 2; = 0, find the value of M
(1).
x²+(2m-1)x+m²=0
There are two real roots
So △ = (2m-1) & sup2; - 4m & sup2; = 1-4m ≥ 0 m ≤ 1 / 4
(2).
Because X1 & sup2; - x2 & sup2; = 0
So X1 = - x2 or X1 = x2
So there are two opposite numbers, namely X1 = - x2
Then X1 + x2 = 1-2m = 0
M = 1 / 2
The general formula of quadratic equation (X-2) (3x-5) = 1 is_____
In addition, if the quadratic equation of one variable X & # 178; - 3x + 2 = 0 is X &; X &;, then x &; + X &=——
The other is to solve the equation x & # 178; + 3x = - 1 with the root formula, and first obtain B & # 178; - 4ac = - -, then x & # 8321; = - -, X & # 8322=————
(1) Fill in: 3x & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\; + 3x + 1 = 0A = 1 b = 3
The general formula is ax ~ + BX + C = 0 (a is not equal to 0)
（X-2）（3X-5）=1
3x~-5x-6x+10-1=0
3x~-11x+9=o
x~- 11 =3=0
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Question: Well, there are two more questions....
If inequality x ^ 2 + 2XY
Let y = KX, (k 〉 0) be substituted by the original inequality,
It is reduced to (2k + 1) x ^ 2 = 1 / 2 (where it can not be strictly greater than), and the minimum value of M can be taken as 1 / 2
It's much easier to solve the problem of limit,
The minimum value of M is 1
Known 2007x + 2008y = 2006, 2008x + 2007y = 2009, calculate (X-Y)
Known (equations)
2007X+2008Y=2006
2008X+2007Y=2009
Find the value of (X-Y) 4 power - (x + y) 2008 Power
Solution 1: 2007x + 2008y = 2006
2:2008X+2007Y=2009
Let formula 1-2 = Y-X = 3
The formula 2 + 1 = 4015x + 4015y = 4015
=X+Y=1
So x = - 1, y = 2
So (X-Y) ^ 4 = 81
(X+Y)^2008=1
So (X-Y) 4 power - (x + y) 2008 power = 81-1 = 80
Is there such a nonnegative integer m that the square of the quadratic equation m with respect to x multiplied by the square of x minus (2m minus 1) multiplied by x plus 1 = 0 has two real roots
In, please write the value of M, if not, please explain the reason
Is there such a nonnegative integer m that the square of the quadratic equation m with respect to x multiplied by the square of x minus (2m minus 1) multiplied by x plus 1 = 0 has two real roots
m^2x^2-(2m-1)x+1=0
When m = 0, the equation is a linear equation with one variable, and x = - 1 satisfies the requirement of the topic
When m ≠ 0,
Discriminant = 4m ^ 2 + 1-4m-4m ^ 2 > = 0
That is m
(X-5) (X-6) = X-5 solution of quadratic equation with one variable
（X-5）（X-6）=X-5
（X-5）（X-6）-(X-5)=0
(x-5)(x-6-1)=0
(x-5)(x-7)=0
x1=5 x2=7
Given that the real numbers x and y satisfy x2 + y2 = 2, then the maximum value of (y + 2) / (x + 2) is
I want the solution, I know the answer
Let k = (y + 2) / (x + 2)
y=kx+(2k-2)
Substituting x ^ 2 + y ^ 2 = 2
So x ^ 2 + K ^ 2x ^ 2 + 2K (2k-2) x + (2k-2) ^ 2 = 2
So (1 + K ^ 2) x ^ 2 + 2K (2k-2) x + (2k-2) ^ 2-2 = 0
This equation about X has a solution
So the discriminant is greater than or equal to 0
4k^2(2k-2)^2-4(1+k^2)[(2k-2)^2-2]>=0
4k^2(2k-2)^2-4(1+k^2)(2k-2)^2+8(1+k^2)>=0
(2k-2)^2(4k^2-4-4k^2)+8(1+k^2)>=0
-4(2k-2)^2+8(1+k^2)>=0
-k^2+4k-1>=0
k^2-4k+1
Given 1 + X + x ^ 2 = 0, what is the value of polynomial 1 + X + x ^ 2 + x ^ 3 + x ^ 4 +... + x ^ 2008 + x ^ 2009
1 + X + x ^ 2 + x ^ 3 + x ^ 4 +... + x ^ 2008 + x ^ 2009 = 0 + x ^ 3 + x ^ 4 +... + x ^ 2008 + x ^ 2009 = x ^ 3 (1 + X + x ^ 2) + x ^ 6 +... + x ^ 2008 + x ^ 2009 = x ^ 6 +... + x ^ 2008 + x ^ 2009 = x ^ 6 (1 + X + x ^ 2) + x ^ 9 +... + x ^ 2008 + x ^ 2009 = x ^ 9 +... + x ^ 2008 + x ^ 2009 so repeatedly extract the common factor 1 + X + X + x ^ 2
His answer is correct
There is no meaning in the real number field, because 1 + X + x ^ 2 = (x + 1 / 2) ^ 2 + 3 / 4 > 0
In the complex field
The formula to be solved = (1 + X + x ^ 2) + x ^ 3 * (1 + X + x ^ 2) + x ^ 6 * (1 + X + x ^ 2) + +x^2007*(1+x+x^2)=0