If real numbers x and y satisfy the condition y = x ^ 2, find the value of log2 (4 ^ x + 4 ^ y) h

If real numbers x and y satisfy the condition y = x ^ 2, find the value of log2 (4 ^ x + 4 ^ y) h

4 ^ x > 0,4 ^ y > 0 so 4 ^ x + 4 ^ y > = 2 √ (4 ^ x * 4 ^ y) = 2 √ 4 ^ (x + y) = 2 √ [2 ^ (x + y)] ^ 2 = 2 * 2 ^ (x + y) = 2 ^ (x + y + 1) = 2 ^ (x ^ 2 + X + 1) x ^ 2 + X + 1 = (x + 1 / 2) ^ 2 + 3 / 4 > = 3 / 4 so 2 ^ (x ^ 2 + X + 1) > = 2 ^ (3 / 4) so log2 (4 ^ x + 4 ^ y) > = log2 2 ^ (3 / 4) = 3 / 4, that is log2 (4 ^ x + 4 ^ y) > = 3 / 4

If y = log2 (x + 1-A / x) increases on [1, + infinity], then the value range of real number a is

x+1-a/x>0
If y = log2x increases monotonically over the positive real number field, it is equivalent to y = x-a / X increasing in [1, + infinity]. If the derivative of y = 1 + A / x ^ 2 > 0, then a > = - x ^ 2 and - x ^ 2

The function y = f (x) (x ≠ 0) is an odd function and an increasing function when x ∈ (0, positive infinity). If f (1) = 0, the inequality f [x (x-1 / 2)]

Solution:
① When x (x-1 / 2) > 0, it is an increasing function
f[x(x-1/2)]

Solving inequality 2log2 (2 ^ x + 1)

Obviously, 2 ^ (x + log2 (5)) = 2 ^ x * 2 ^ (log2 (5)) = 5 * 2 ^ x, and 2log2 (2 ^ x + 1) = log2 (2 ^ x + 1) ^ 2, the base of two logarithms is 2, so (2 ^ x + 1) ^ 2 ≤ 5 * 2 ^ X - 1, expansion and simplification can get, (2 ^ x) ^ 2 - 3 * 2 ^ x + 2 ≤ 0, that is, (2 ^ X - 1) (2 ^ X - 2) ≤ 0, so 1 ≤ 2 ^ x

On the inequality k * (4 Λ x) - 2 Λ (x + 1) + 6K < 0 of X, if the equality holds for all x ∈ {x | 1 < x < log2 (3)), find the value range of real number K
Let 2 ^ x = t be (2,3)
So KT ^ 2-2t + 6K

k*(4∧x)-2∧(x+1)+6k＜0
That is, K (2 ^ x) ^ 2-2 * (2 ^ x) + 6K

If x belongs to R, the inequality x ^ 2 * log2 [4 (1 + a) / a] + 2xlog2 [2A / (a + 1)] + log2 [(a + 1) ^ 2 / 4 * a ^ 2] > 0 holds, and the value range of real number a is obtained
Try to have a process, but there must be an answer

Log2 [4 (1 + a) / a] (X & # 178;) * [2A / (a + 1)] ^ 2x * [(a + 1) ^ 2 / 4 * a ^ 2]] is greater than 0
4 (1 + a) / a] (X & # 178;) * [2A / (a + 1)] ^ 2x * [(a + 1) ^ 2 / 4 * a ^ 2] greater than 1
A belongs to [- 1 / 3,1]

It is known that a ∩ B can be obtained from the set a = {(x, y) │ 3x-2y = 11}, B = {(x, y) │ 2x + 3Y = 16}

3x-2y=11
2x+3y=16
Just solve the equations, x = 5, y = 2
So a ∩ B = (5,2)

Given the set a = {y | y = x2-2x, X ∈ r}, B = {Z | z = x2 + 6x + 8, X ∈ r}, find the intersection of a and B
The process should be detailed

A = {y | y = x ^ 2-2x, X ∈ r}, a = {y | y = x (X-2)} B = {Z | z = x ^ 2 + 6x + 8, X ∈ r} = {Z | z = (X-2) (x-4)} A and B = > (X-2) (x-4) = x (X-2) x ^ 2 + 6x + 8 = x ^ 2-2x8x = - 1When x = - 1y = - 1 (- 1-2) = 3A and B = {3}#

Given a = (Y / y = - x2-6x + 8, X belongs to R), B = (y = - x2 + 2x + 7, X belongs to R), find the union of a and B, the intersection of a and B
Urgent need

This is to find the range. The range of a is negative infinity to 17 left open and right, and that of B is negative infinity to 8 left open and right. Then the answer comes out. The union of a and B is negative infinity to 17 left open and right closed, and the intersection is negative infinity to 8 left open and right closed

Given a = {y = x ^ 2-2x, X ∈ r}, B = {z = x ^ 2 + 6x + 8, X ∈ r} to find a ∩ B

A={y=x^2-2x,x∈R},B={z=x^2+6x+8,x∈R}
for A∩B =>
x^2-2x = x^2+6x+8
8x = -8
x = -1
for x=-1
y=1+2 = 3
A∩B = {3} #