If the solution of the equation MX-1 = 2x of X is a positive real number, then the value range of M

If the solution of the equation MX-1 = 2x of X is a positive real number, then the value range of M

mx-1=2x
mx-2x=1
(m-2)x=1
x=1/(m-2)
The solution is a positive real number
1/(m-2)＞0
m-2＞0
m＞2
mx-1=2x
mx-2x=1
(m-2)x=1
Because (m-2) x is not zero
So m-2 is not zero
So m is not equal to 2
If the solution of the equation MX-1 = 2x of X is a positive real number, then the value range of M is ()
A. m≥2B. m≤2C. m＞2D. m＜2
From MX-1 = 2x, term shifting and merging, we get (m-2) x = 1, ∵ x = 1m − 2. The solution of the equation MX-1 = 2x is a positive real number, ∵ 1m − 2 > 0, and the solution is m > 2
If the solution of the equation MX-1 = 2x about X is a positive real number, then the value range of M is?
mx-2x=1
x(m-2)=1
x=1/(m-2)
Because x > 0 and m-2 is not equal to 0
So x > 2
If the solution of the integral equation MX-1 = 2x of X is a positive real number, what is the value range of M?
mx-1=2x
mx-2x=1
(m-2)x=1
Because the solution of X is a positive real number
So x = 1 / (m-2) > 0
So m-2 > 0
M>2
mx-1=2x;
(m-2)x=1;
x=1/(m-2)＞0;
m＞2;
The decreasing interval of function y = x2-6x is ()
A. （-∞，2]B. [2，+∞）C. （-∞，3]D. [3，+∞）
The equation of symmetry axis of ∵ function y = x2-6x is x = 3, and the corresponding image is a parabola with the opening upward. As shown in the figure, the subtraction interval of ∵ function y = x2-6x is (- ∞, 3]
Given that the function f (x) = √ (4x-x ^ 2) - X-B has only one zero point, then the value range of B
F (x) = √ (4x-x ^ 2) - X-B has only one zero point, that is, let Y1 = root sign (4x-x ^ 2), that is, let x ^ 2-4x + y ^ 2 = 0, (X-2) ^ 2 + y ^ 2 = 0, which means the upper half of a circle with radius 2 and center (2,0). Let y2 = x + B, which means a straight line, and f (x) = y1-y2 has only one zero point, which means that Y1 and Y2 have only one intersection point
The monotone decreasing interval of the function y = 32 − 3x2 is______ .
The function y = 32 − 3x2 is a composite function, and the domain of definition is R. the outer function is y = 3T, and the inner function is t = 2-3x2. Because the outer function y = 3T is an increasing function, and the inner function T = x2 + 2x is an increasing function on (- ∞, 0) and a decreasing function on (0, + ∞), the monotonic decreasing interval of the composite function y = 32 − 3x2 is: (0, + ∞), so the answer is: (0, + ∞) & nbsp; Note: [0, + ∞) & nbsp; can also be used
If the function f (x) = ax-x-a (a > 0 and a ≠ 1) has two zeros, then the value range of real number a is ()
A. 0＜a＜1B. 0＜a＜12C. a＞2D. a＞1
If the function f (x) = ax-x-a (a > 0 and a ≠ 1) has two zeros, then the function y = ax and y = x + a have two intersections. When 0 < a < 1, the function y = ax and y = x + a have only one intersection, which does not meet the condition. When a > 1, the function y = ax and y = x + a have two intersections, as shown in the figure: therefore, the value range of real number a is a > 1. Therefore, select D
Monotone decreasing interval of function y = 3 ^ (2x ^ 2-3x + 6)
People who know speed
3>1
So 3 ^ x is an increasing function
So the minus interval of Y is the minus interval of index
2x^2-3x+6=2(x-3/4)^2+39/8
The opening is up, so X
X1. Same increase different decrease. Therefore, we only need to reduce the quadratic polynomial! So x
Let f (x) = P (x-1 / x) - INX, G (x) = 2E / X (P is a real number, e is the base of natural logarithm)
If there is at least one x 0 (0 is a subscript) on [1, e], such that f (x 0) > G (x 0) holds, the value range of P is obtained
The trouble must be solved before tonight, I will add it
f(x)=p(x- 1/x)-2lnx
Finding the no proposition is to find the complement on [1, e] if f (x) ≤ g (x)
Let H (x) = P (x-1 / x) - inx-2e / X not know what your f (x) is
X-1 / X or (x-1) / x, let's provide an idea and do it by ourselves