F (x) = 2 radical 3sin (ω x + π / 3) (ω > 0), f (x + θ) is an even function with period 2 π (1) Find the value of ω and θ (2) When x ∈ [- π / 2, π / 12], find the range of F (x)

F (x + θ) = 2 radical 3sin (ω x + ω θ + π / 3), so w = 1, ω θ + π / 3 = π / 2 + K π, θ = π / 6 + K π, less conditions, K can not be found

When x belongs to [0, Pai / 2), f (x) = radical 3tanx-1, find f (8 Pai / 3)

f(8π/3)=f(8π/3-2π)=f(2π/3)=f(-π+2π/3)=f(-π/3)=f(π/3)=√3tan(π/3)-1=(√3)²-1=3-1=2

Simplification of equation (square of (X-2) + square of (x + 2) + square of (x + 2) + square of y = 10

(the square of (X-2) under the root sign + the square of Y) = the square of (x + 2) + the square of Y under the root sign
The square of (x + 2) + the square of y = 5 + 2 / 5x
The square of (x + 2) + the square of y = 25 + 4x + 0.16x
84x squared + y squared = 21

If x is less than 2, the square + absolute value 3-x absolute value of root sign (X-2) is simplified=

√(x-2)^2+|3-x|
=2-x+3-x
=5-2x

Let a be the sum of the absolute values of all the roots of the equation x ^ 2 - root number 2003x-520 = 0, then a ^ 2 =? 0?

Let a be the sum of the absolute values of all the roots of the equation x square root sign 2005x-520 = 0, and find the value of the square of A

Let the two equations be X1 and X2 respectively
X^2—√2005X—520=0
According to Weida's theorem:
X1+X2=√2005,x1x2=-520
Because: a = | x1 | + | x2|
Then: A ^ 2 = (| x1 | + | x2 |) ^ 2
=|x1|^2+|x2|^2+2|x1||x2|
=x1^2+x2^2+2|x1x2|
=(x1^2+x2^2+2x1x2)-2x1x2+2|x1x2|
=(x1+x2)^2-2x1x2+2|x1x2|
=2005-2(-520)+2|-520|
=2005+1040+1040
=4085

What is the curve represented by 1-x ^ 2 under the equation y = - radical?

If y = - √ 1-x ^ 2, then y ^ 2 = 1-x ^ 2, that is, x ^ 2 + y ^ 2 = 1
And here y ≤ 0, so you get the lower half of the image x ^ 2 + y ^ 2 = 1

The curve represented by 2 (x-1) ^ 2 + 2 (Y-1) ^ 2 = x + Y-2 under the root sign of the equation is

√{2(x-1)^2+2(y-1)^2} = x+y-2
x+y-2≥0,x+y≥2
The original formula is: √ {2 (x-1) ^ 2 + 2 (Y-1) ^ 2} = (x-1) + (Y-1)
Square of both sides: 2 (x-1) ^ 2 + 2 (Y-1) ^ 2 = (x-1) ^ 2 + (Y-1) ^ 2 + 2 (x-1) (Y-1)
Simplification:
(x-1)^2+(y-1)^2 - 2(x-1)(y-1) = 0
(x-1-y+1)^2 = 0
x-y=0
Y=x
And: x + y ≥ 2
ν denotes that the ray y = x, and the definition domain x ∈ [1, + ∞)

Draw the curve represented by the equation X-1 = radical 1-y ^ 2 It's best to have a picture

X-1 = radical 1-y ^ 2
It can be changed to (x-1) ^ 2 = 1-y ^ 2
That is (x-1) ^ 2 + y ^ 2 = 1
So the graph is a circle with (1,0) as the center and 1 as the radius

What is the curve represented by 3 (x + 1) ^ 2 + 3 (y + 1) ^ 2 = | x + Y-2 | under the root sign of the equation? The explanation is the distance from (x, y) to (- 1, - 1)

Change the equation to √ [(x + 1) ^ 2 + (y + 1) ^ 2] = | x + Y-2 | / √ 2 * √ (2 / 3),
It means that the ratio of the distance from point to point (- 1, - 1) with coordinates (x, y) to the distance from the straight line x + Y-2 = 0 is √ (2 / 3),
Because √ (2 / 3) = √ 6 / 3 < 1,
So it's an ellipse

F (x) = 2 radical 3sin (ω x + π / 3) (ω > 0), f (x + θ) is an even function with period 2 π (1) Find the value of ω and θ (2) When x ∈ [- π / 2, π / 12], find the range of F (x)