Simplify - 7x square + 6 (x square - 5xy) - 2 (3xy square - x square)

Simplify - 7x square + 6 (x square - 5xy) - 2 (3xy square - x square)

-7xsquare + 6 (xsquare-5xy) - 2 (3xy-xsquare)
=-7x square + 6x square - 30xy-6xy square + 2x square
=X square - 30xy-6xy square

What is the sum of (square of X - 2x + 5) (square of X + 2x-5)

(square of X - 2x + 5) (square of X + 2x-5)
=(x^2-2x)(x^2+2x)-25
=x^4-4x^2-25

Given that a and B are straight lines L; 3x + 4Y + 2 = 0 intersect the focus of circle x ^ 2 + y ^ 2 + 4Y = 0, what is the equation of the vertical bisector of line segments a and B

The vertical bisectors of line segments a and B must pass through the center of the circle and be perpendicular to the original line
So k = 4 / 3. Center coordinates (0, - 2), so the linear equation is:
Y + 2 = 4 / 3x, that is 4x-3y-6 = 0

How to judge that the axis of symmetry of F (x + a) = f (x-a) is x = a?
How to judge?

The period f (x + a) = f (x-a) is t = 2A
If f (a + x) = f (A-X), the axis is x = a
[(a+x)+(a-x)]/2=a

The value range of X which makes √ x + 1, 1 / 2 of √ x, (x + 4) ^ 0 meaningful is

The three formulas √ x + 1, √ 1 / 2 of X, (x + 4) ^ 0 are all meaningful
Namely
x+1≥0 x≠0 x+4≠0
x≥-1,x≠0 x≠-4
So the value range of X is x ≥ - 1 and X ≠ 0
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If the equation k2x2 + 2 (k-1) x + 1 = 0 of X has two real roots, then the value range of K is ()
A. K < 12b. K ≤ 12C. K < 12 and K ≠ 0d. K ≤ 12 and K ≠ 0

(1) ∵ the quadratic equation of one variable k2x2 + 2 (k-1) x + 1 = 0 has two real roots, ∵ (?) = [2 (k-1)] 2-4k2 ≥ 0 and K2 ≠ 0, and the solution is k ≤ 12 and K ≠ 0

3（x+2）-2（x-3）=16 2.5（5-x）=3（x+4）-3

3（x+2）-2（x-3）=16
3X+6-2X+6=16,
X=4
2.5（5-x）=3（x+4）-3
25-5X=3X+12-3
-8X=-16
X=2.

For a project, it takes a day for Party A to complete it alone, and it takes B days for Party B to complete it alone. Both a and B are natural numbers. Now Party B works for three days first, and Party A and Party B work together for one day to complete it, then the value of a + B is equal to______ Or______ ．

1A + 1b × (1 + 3) = 1, ∵ a, B are natural numbers, ∵ a = 3, B = 6; or a = 5, B = 5; or a = 2, B = 8; the value of a + B is 9 or 10

8x-4 = 3x + 6-1

8x-4=3x+6-1
8x-3x=6-1+4=9
5x=9
x=9/5

F (x) = (1-cos2x) / X for derivative f '(0) =?
F (x) = (1-cos2x) / X for derivative f '(0) =?
The answer is f '(0) = 2

f'(x)=[2xsin2x-(1-cos2x) ]/x²
Because the denominator is x, when x = 0, it doesn't make sense
therefore
The derivative does not exist