The known set a = {0, a}, B = {B | B ^ 2-3b

The known set a = {0, a}, B = {B | B ^ 2-3b

B=｛b|b^2-3b

What is the solution set of the inequality x | x | + 3 | x | + 2 > 0

When x is greater than or equal to 0, when x is less than 0, X | x | + 3 | x | + 2 = - XX + 3x + 2 is greater than 0, the solution is (3-radical 17) / 2 to 0, and the solution is (3-radical 17) / 2 to positive infinity

Solve the inequality a (x-1) / X-2 > 1 about X. thank you,

The inequality a (x-1) / X-2 ＞ 1 multiplies x on both sides to get a (x-1) - 2x ＞ x ax-a-3x ＞ 0 x (A-3) ＞ a x ＞ A / (A-3). Please

The function f (x) is known to hold f (AB) = f (a) + F (b) for any real number a and B
1) Finding the value of F (1) and f (0)
2) If f (2) = P, f (3) = q (P, q are constant), find the value of F (36)
3) Prove f (1 / x) = - f (x)

(1) Let a = b = 1F (1 × 1) = f (1) + F (1) f (1) = f (1) + F (1) so f (1) = 0 let a = b = 0f (0 × 0) = f (0) + F (0) f (0) = f (0) + F (0) so f (0) = 0 (2) f (36) = f (2 × 18) = f (2) + F (18) = P + F (2 × 9) = P + F (2) + F (9) = P + P + F (3 × 3) = P + P + F (3) + F (3) = P + P + Q + q = 2 (P + Q) (3) let a

Please note that when a and B take any value, the value of polynomial a square + b square - 2A + 6B + 12 is always positive

A square + b square - 2A + 6B + 12
=(A-1) square + (B-3) square + 12-1-9
=(A-1) square + (B-3) + 2 > 0
The value of polynomial a square + b square - 2A + 6B + 12 is always positive

No matter what value a and B take, the value of polynomial A & # 178; B & # 178; + 4B & # 178; - 6ab-4b + 12 is not less than 2

Given that the real numbers a and B satisfy the square of A-3 + B + 4B + 4 = 0, then what is the 2010 power of (a + b) / the 2009 power of (a + b)

The square of A-3 + B + 4B + 4 = 0, that is, A-3 + (B + 2) ^ 2 = 0, the solution is a = 3, B = - 2, a + B = 1
therefore
2010 power of (a + b) / 2009 power of (a + b) = 1 / 1 = 1

It is known that the area of a rectangle is 4a2-2ab + 14b2, and the length of one side is 4a-b, then the perimeter of the rectangle is______ ．

Then the circumference of the rectangle = [14 (4a-b) + (4a-b)] × 2 = [a-14b + 4a-b] × 2 = [5a-54b] × 2 = 10a-52b

The area of a rectangle is 4A ^ 2-2ab + 1 / 4B ^ 2. It is known that the length of one side is 4a-b. the perimeter of the rectangle is calculated

4a^2-2ab+(1/4)b^2
=4a^2-ab-ab+(1/4)b^2
=a*(4a-b)-(b/4)*(4a-b)
=(4a-b)*(a-b/4)
So the other side is A-B / 4
Perimeter = (a-b / 4 + 4a-b) × 2
=10a-5b/2

The perimeter of a rectangle is 4A + 4B. If the length of one side of the rectangle is expressed by a, then the area of the rectangle is?