Given that X and y are positive integers, and (x - √ (3Y)) & sup2; = 21-12 √ 3, find the value of X-Y

Given that X and y are positive integers, and (x - √ (3Y)) & sup2; = 21-12 √ 3, find the value of X-Y

(x-√(3y))² = 21-12√3 = (2√3 - 3)²
x-√(3y) = +(-)(2√3 - 3)= +(-)(3 - 2√3)
Because X and y are positive integers, so
(1) When X - √ (3Y) = (3 - 2 √ 3), the,
x = 3
y = 4
x-y = -1
(2) When X - √ (3Y) = - (3 - 2 √ 3), X and y cannot be positive integers, so there is no solution
So, X - y = - 1

What is the calculation result of (x-2y) 2 + (x + 2Y) (x-2y) - (x-3y) (X-Y)? X is equal to - 4 and Y is equal to 0.5
2 after the first bracket means square!

(x-2y) ^ 2 + (x + 2Y) (x-2y) - (x-3y) (X-Y) = (x-2y) [(x-2y) + (x + 2Y)] - (x-3y) (X-Y) = (x-2y) 2x - (x-3y) (X-Y) = 2x ^ 2 - 4xy - x ^ 2 + 4xy - 3Y ^ 2 = x ^ 2 - 3Y ^ 2 after substitution = (- 4) ^ 2 - 3 * 0.5 ^ 2 = 16 - 3 / 4 = 15 1 / 4 = 15.25

Set a = {(x, y) | y = 2 x power, X ∈ r} set B = {(x, y) | y = x 2 power, X ∈ r}, P = a ∩ B, then the number of elements in set P is several

Set a = {(x, y) | y = 2 ^ x, X ∈ r}
Set B = {(x, y) | y = x & # 178;, X ∈ r}
There are only two p = a ∩ B = {(2,4), (4,16)}

The interval of real number solution of function f (x) = X3 + x-3 is ()
A. 〔0，1〕B. 〔1，2〕C. 〔2，3〕D. 〔3，4〕

∵ f '(x) = 3x2 + 1 ≥ 0 ∵ the function f (x) = X3 + x-3 is a monotone increasing function on R ∵ f (1) = 1 + 1-3 = - 1 < 0, f (2) = 8 + 2-3 = 7 > 0 ∵ the interval of real number solution of function f (x) = X3 + x-3 is (1,2), so B is chosen

According to the dichotomy, the interval where the real solution of the function f (x) = X5 + x-3 falls is ()
A. [0，1]B. [1，2]C. [2，3]D. [3，4]

Let f (x) = X5 + x-3, substitute x = 0, 1, 2, 3, 4, if f (a) · f (b) < 0, then the zero point is in (a, b), so f (1) < 0, f (2) > 0, so in (1, 2), so select B

If the function f (x) = AX2 + 2 (A-3) x + 1 is a decreasing function in the interval (- 2, + ∞), then the value range of real number a is?

Discussion
When a = 0
f=-6x+1
Meet the requirements
When a ≠ 0
According to the meaning of the title
f‘=2ax+2a-6

For y = 3cos (2x + π / 3) + 1, the domain of definition, range of values, monotone interval, parity, period, maximum value, axis of symmetry, center of symmetry,

What process can this have? This is the most basic thing. We need to see the definition domain at a glance: R range: [- 2,4] monotone increasing interval (- 2 π / 3 + K π, - π / 6 + K π), K ∈ Z single decreasing interval (- π / 6 + K π, π / 3 + K π) parity: non odd non even function minimum positive period T = π, maximum value is 4, minimum value is - 2 symmetry axis X = - π

Y = 2sinx (2x - π / 3) + 1 to find the maximum and minimum value of the domain of definition

The definition domain is r, the range is [- 1,3], the maximum is 3 (sin (2x - π / 3) = 1), and the minimum is - 1 (sin (2x - π / 3) = 1). The symmetric axis equation x = k π / 2 + 5 π / 12 (let 2x - π / 3 = k π + π / 2). The symmetric center coordinates (K π / 2 + π / 6,1). The monotonic increasing interval is [K π - π / 12, K π + 5 π / 12]

Given the function F X = 3cos (2x + π / 4), find the monotone decreasing interval

F (x) = 3cos (2x + π / 4) f '(x) = - 6sin (2x + π / 4) let: F (x) < 0, that is: - 6sin (2x + π / 4) < 0, then: sin (2x + π / 4) > 0, and the solution: K π - π / 8 < x < K π + 3 π / 8, k = 0, ± 1, ± 2, ± 3 The monotone decreasing interval of the function is: X ∈ (K π - π / 8, K π + 3 π / 8), k =

Prove: (Tan ^ 2) x - (sin ^ 2) x = (Tan ^ 2) x (sin ^ 2) X

Notice thattan²(x) = sin²(x) / cos²(x) .Using this on the right-hand side of the equation and gathering into a common fraction,we gettan²(x) - sin²(x) = (sin²(x) / cos&su...