Known about X, y equation 7 (A-3) x ^ | a | - 2 + (B + 2) y ^ | 2B | - 3 = - 5 is a binary linear equation, find the value of (a-b) ^ A + B!

Known about X, y equation 7 (A-3) x ^ | a | - 2 + (B + 2) y ^ | 2B | - 3 = - 5 is a binary linear equation, find the value of (a-b) ^ A + B!

a-3≠0
|a|-2=1
b+2≠0
|2b|-3=1
The solution is a = - 3, B = 2
(a-b)^a+b=(-3-2)^(-3)+2=-1/125+2=249/125

Is the square of 2m-1 * x + 3x-5 = 0 (M is a constant) a quadratic equation of two variables

No

Given that the minimum value of quadratic function f (x) = (LGA) x ^ 2 + 2x + 4lga is 3, find the value of real number a

Let m = LGA, then f (x) = MX ^ 2 + 2x + 4m
① When m = 0, i.e. a = 1, then f (x) = 2x and the function has no minimum value
② , m ≠ 0, we can see that f (x) is a quadratic function, and because there is a minimum value for all x, the opening of the image should be upward,
The results show that: when m ＞ 0, f (min) = (4ac-b ^ 2) / 4A = (16m ^ 2-4) / 4m = 4m ^ 2-1, the minimum value is 3, 4m ^ 2-1 = 3,
We obtain M = 1 or M = - 1 (rounding off, because m ＞ 0). We obtain LGA = 1 and a = 10

Find the maximum value of the function y = 1-2x-3 / X (x is greater than 0)

Y = 1-2x-3 / x, find the derivative, get y '= - 2 + 3 / x ^ 2 = (3-2x ^ 2) / x ^ 2 > 0, get - √ 6 / 2

(1) Find the maximum value of the function y = x (a-2x) (x > 0, a is a constant greater than 2x); (2) let x > 1, find the maximum value of the function y = (x + 5) (x + 2) x + 1

(1) ∵ x ＞ 0, a ＞ 2x, ∵ y = x (a-2x) = 12 × 2x (a-2x) ≤ 12 × [2x + (a − 2x) 2] 2 = A28, if and only if x = A4, take the equal sign, so the maximum value of the function is A28. (2) ∵ x ＞ - 1, ∵ x + 1 ＞ 0, let x + 1 = Z ＞ 0, then x = Z-1, ∵ y = (Z + 4) (Z + 1) z = Z2 + 5Z + 4Z = Z + 4Z + 5 ≥ 2Z + 5 =

The maximum value of the function y = - 2x Λ & #178; + 4x-5 is, which is correct

Given the function f (x) = x ^ 2-2x, G (x) = x ^ 2-x ([2,4]). Find the monotone interval of F (x), G (x). Find the minimum

The minimum value of the two functions is 0,2

Given the function f (x) = x2-2x, G (x) = x2-2x, X ∈ [2,4]. (1) find the monotone interval of F (x), G (x); (2) find the minimum of F (x), G (x)

(1) The axis of symmetry of the function f (x) = x2-2x is x = 1; the function f (x) decreases monotonically on (- ∞, 1) and increases monotonically on (1, + ∞); the function g (x) increases monotonically on [2,4]; (2) it is known from (1) that the minimum value of function f (x) is - 1 when x = 1, and the minimum value of function g (x) is 0 when x = 2

Given the function f (x) = x & sup2; - 2x g (x) = x & sup2; - 2x (x ∈ [2,4]) 1, it is necessary to find the monotone interval 2 of F (x) g (x) and the minimum value of F (x) g (x)
I can't do it
I don't know what is the standard answer
Please give the most standard answer

f(x)=x^2-2x=(x-1)^2-1
When x = 1, f (x) is the minimum, the minimum = - 1
When + infinity > x > 1, f (x) increases monotonically
When infinity

The monotone decreasing interval of function y = - x ^ 2 + 2x-a is
Need detailed process, otherwise I don't understand

y=-x²+2x-a
It is a parabola with the opening downward and the axis of symmetry x = 1
The right side of the axis of symmetry is decreasing
So, the monotone decreasing interval is (1, + ∞)