Given the - 2nd power of x-x = a, find the square of X + the - 2nd power of X

Given the - 2nd power of x-x = a, find the square of X + the - 2nd power of X

x-1/x=a
Two sides square
（x-1/x)^2=a^2
x^2+1/x^2-2=a^2
x^2+1/x^2=a^2+2
It is known that the - 1 power of x-x = a,
Two sides square X & # 178; - 2 + x ^ (- 2) = A & # 178;
So x & # 178; + x ^ (- 2) = A & # 178; + 2: How did X & sup2; - 2 + x ^ (- 2) = A & sup2; come from?
The increasing interval of the function y = log2 (- the square of X - 2x + 8) is
y=log2(-x^2-2x+ 8）
f(x)=-x^2-2x+ 8=-(x^2+2x+1)+9=-(x+1)^2+9
The function increases monotonically in (negative infinity, - 1) and decreases monotonically in [- 1, positive infinity)
F (x) = log2 (x) is an increasing function
So the monotone increasing interval of the function y = log2 (- x ^ 2-2x + 8) is: (negative infinity, - 1]
-x^2-2x+8>0
x^2+2x-8
Given the power function f (x) = x (2-k) (1 + k), K ∈ Z, and f (x) monotonically increases on (0, + ∞). (1) find the value of real number k, and write the corresponding analytic expression of function f (x); (2) if f (x) = 2F (x) - 4x + 3 is not monotone in the interval [2a, a + 1], find the value range of real number a; (3) try to judge whether there is a positive number Q, so that the function g (x) = 1-qf (x) + (2q-1) X The value range on the interval [- 1, 2] is [− 4178]. If it exists, find out the value of Q; if it does not exist, explain the reason
(1) From the meaning of (2-k) (1 + k) > 0, the solution is: - 1 < K < 2 (2) k ∈ Z  k = 0 or K = 1 (3 points) are substituted into the original function, f (x) = x2 (4 points) (2) f (x) = 2x2-4x + 3 If the function is not monotone, then 2A < 1 < A + 1, then 0 < a < 1
The monotone increasing interval of the function y = log 1 / 2 (- xsquare - 2x + 24) is
Monotone increasing interval of function y = log 1 / 2 (- xsquare - 2x + 24)
That is the monotone decreasing interval of the function - x square - 2x + 24
And - x squared - 2x + 24 = - (x + 1) squared + 25
The minus interval is (- 1, + ∞)
But the true number must be greater than 0
Namely
-X square - 2x + 24 > 0
X square + 2x-24
If the power function f (x) = k * x ^ α is known and the image passes through a point (1 / 2, √ 2 / 2), then K + α=
Substitution point: √ 2 / 2 = k * (1 / 2) ^ α
The exponent on the left is 1 / 2, so α = 1 / 2 on the right
√2/2=k*(1/2)^（1/2）=k*（√2/2）
So k = 1
K + α = 3 / 2
The monotone increasing interval of function y = log2 (x2-2x-3) is______ .
From x2-2x-3 ＞ 0, we get x ＜ - 1 or X ＞ 3, so the domain of definition of function is (- ∞, - 1) ∪ (3, + ∞). Because y = log2u increases, u = x2-2x-3 increases on (3, + ∞), so y = log2 (x2 − 2x − 3) increases monotonically on (3, + ∞), so the monotonic increasing interval of function y = log2 (x2 − 2x − 3) is (3, + ∞), so the answer is: (3, + ∞)
Let the power function f (x) = (A-1) x ^ k pass (root 2,2) to find a, K
Because it's a power function, A-1 = 1
A=2
Bring in (radical 2,2)
2 = (radical 2) ^ k
K=2
The increasing interval of the function y = log2 (the square of 2x-x) is? Where 2 is the base,
Log based on two is itself an increasing function. If you want to get the increase and decrease of the whole function, it only depends on the formula in brackets. If the formula in brackets is increasing, the whole function will increase. Otherwise, it will decrease. The whole problem is to ask you to find the increasing interval of (2x-x Square) and calculate it by yourself
Firstly, if 2x-x2 > 0, then x0 is obtained. The axis of symmetry of y = 2x-x2 is x = 1, so the increasing interval of y = 2x-x2 is x > 1,
That is to say, the increasing interval of log2 (2x-x2) is x > 1.
If the image of power function y = f (x) passes through points (9, 13), then the value of F (25) is______ .
The image of ∵ power function y = f (x) passes through points (9, 13). Let f (x) = x α, α be constant, ∵ 9 α = 13, ∵ α = - 12, so f (x) = x − 12, ∵ f (25) = (25) − 12 = 15, so the answer is: 15
The increasing interval of function y = log2 / 1 (x ^ 2 + 2x + 3) is?
X ^ 2 + 2x + 3 is always greater than 0
That is to find the decreasing interval of F = x ^ 2 + 2x + 3
Is negative infinity to - 1
Is the function a logarithm of 1 / (x ^ 2 + 2x + 3) with 2 as the base?
If so, it is the increasing interval of y = - log2 (x ^ 2 + 2x + 3), and∵ log2x is the increasing function, which is the decreasing interval of (x ^ 2 + 2x + 3)
The axis of symmetry of x ^ 2 + 2x + 3 = (x + 1) ^ 2 + 2 is a parabola with the opening upward of x = - 1. When - 1 ≤ x, it is a decreasing function, and when x ≥ - 1, it is an increasing function
The increasing interval of Y is (- ∞, - 1]
Or find the derivative y '= - (2x + 2) / LN2 = - 2 (... Expansion)
Is the base of the logarithm function ^ 2 + 2 / (1 + 3)?
If so, it is the increasing interval of y = - log2 (x ^ 2 + 2x + 3), and∵ log2x is the increasing function, which is the decreasing interval of (x ^ 2 + 2x + 3)
The axis of symmetry of x ^ 2 + 2x + 3 = (x + 1) ^ 2 + 2 is a parabola with the opening upward of x = - 1. When - 1 ≤ x, it is a decreasing function, and when x ≥ - 1, it is an increasing function
The increasing interval of Y is (- ∞, - 1]
Or find the derivative y '= - (2x + 2) / LN2 = - 2 (x + 1) / LN2, when x ≤ - 1, y' ≥ 0, y is an increasing function, and Y is an increasing function