How many groups are set a, B and C equal to (1,2,3,4)

How many groups are set a, B and C equal to (1,2,3,4)

In this paper, a 8746; B 8746874687468746874687468746\8746\\8746\\8746\8746\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\45c, a ∪ B ∪ C = a + B + C-B ∩ C-A ∩ B, a ∪ B ∪ C = a + B + C-A ∩ C-A ∩ B-B ∩ C + a ∩ B ∩ C, too many

If set a = {0, 1, 2, 3}, B = {1, 2, 4}, then set a ∪ B=______ ．

∪ a = {0, 1, 2, 3}, B = {1, 2, 4}, ∪ a ∪ B = {0, 1, 2, 3, 4}; so the answer is: {0, 1, 2, 3, 4}

If set a = {1,2,3,4}, B = {2,4,7,8} C = {0,1,2,3,5}, then set (a ∪ b) ∩ C equals ()
Sorry, C = {0,1,3,4,5}

∵A∪B={1,2,3,4,7,8}
∪ set (a ∪ b) ∩ C = {1,3,4}

Finding the solution set of inequality with absolute value 5-2x greater than 4

| 5-2x |>4
5-2x > 4 or 5-2x 9 or - 2x

It is known that y = f (x) is an odd function defined on R. when x is greater than 0, f (x) = x - | lgx |. When x is less than 0, f (x) is solved analytically
I've calculated three times. The answers are different. I'm a little depressed. How can I define the absolute value? Thank you for your help

Y = f (x) is an odd function defined on R
f(x)=-f(-x)
When X0
f(-x)=-x-|lg(-x)|
Because f (x) = - f (- x)
So f (x) = - (- X - | LG (- x) |) = x + | LG (- x)|

It is known that y = f (x) is an odd function defined on R. when x ≥ 0, f (x) = x ^ 2-2x, then what is the expression of F (x) on R?
How can I simplify my f (x) = x ^ 2-2x (x ≥ 0) f (x) = - x ^ 2-2x (x < 0) y = x (| x | - 2)

F (x) is an odd function defined on R
Then f (0) = 0
x＜0,
-x＞0
So, f (- x) = (- x) ^ 2-2 (- x)
Because it's an odd function
So, f (- x) = - f (x) = - (x ^ 2 + 2x) = - x ^ 2-2x
To sum up, the expression of F (x) on R is: F (x) = x ^ 2-2x (x ≥ 0)
f(x)=-x^2-2x （x＜0）
That is y = x (| x | - 2)
Note that when x is a positive or negative number, the change of X depends on the change of the analytic expression of the function

The smaller area enclosed by y = | x | and circle x2 + y2 = 4 is ()
A. π4B. πC. 3π2D. 3π4

Draw the graph of function in the coordinate system, one is a circle with radius of 2, the other is a broken line, the smaller area is one fourth of the area of the circle, the area is 14 × π × 22 = π, so choose B

The equation x ^ 2 + kx-3 = 0 and x ^ 2 + 4x - (k-1) = 0 have only one same root, and the calculated value and the same root

In x ^ 2 + kx-3 = 0 △ = k ^ 2 + 12 = 0, so this k value has no solution
In x ^ 2 + 4x - (k-1) = 0, △ = 16 + 4 (k-1), because we want a root to be the same, so this △ = 0, solve k = - 3, and then bring the K value to two formulas to make them x ^ 2 + kx-3 = x ^ 2 + 4x - (k-1), then we can calculate x = - 1

When the value of K is what, the equation x ^ 2 + kx-3 = 0 and the equation x ^ 2-4x - (k-1) = 0 have a common root?

From the meaning of the question: x ^ 2 + KX – 3 = 0, x ^ 2 - 4x – (K – 1) = 0, from X2 - 4x – (K – 1) = 0: k = x ^ 2 - 4x + 1 to x ^ 2 + KX – 3 = 0: x ^ 2 + (x ^ 2 - 4x + 1) x – 3 = 0, x ^ 3 - 3x ^ 2 + X – 3 = 0, X – 3 (x ^ 2 + 1) =

So the other root of the equation is K2 + 4x = 4x

Substituting x = 2 into the original equation
4·2²-2k+3=0
k=19/2
The original equation is 8x & # 178; - 19x + 6 = 0
(x-2)(8x-3)=0
x1=2,x2=3/8
The other root is 3 / 8