Given that the absolute value of X-1 + the square of (Y-2) + the root (Z + 3) = 0, find the value of the square of XY + root Z

Given that the absolute value of X-1 + the square of (Y-2) + the root (Z + 3) = 0, find the value of the square of XY + root Z

The absolute value of X-1 + the square of (Y-2) + the root sign (Z + 3) = 0
So X-1 = 0, Y-2 = 0, Z + 3 = 0
x=1,y=2,z=-3
So the original formula = XY + | Z|
=1*2+|-3|
=5

Value range: y = 2Sin (2x + π / 3), X ∈ [- π / 6, π / 6]

y=2sin（2x+π/3）,
∵x∈【-π/6,π/6】
∴2x+π/3∈【0,2π/3】
When 2x + π / 3 = π / 2, there is a maximum value of ymax = 2
When 2x + π / 3 = 0, there is a minimum value of Ymin = 0
Range [0,2]

The main process of finding the value range of function y = (3-x) / 2x + 5 (x ≥ 0)

y=(3-x)/(2x+5)
∵X≥0,
∴①3-X≤3
②2X+5≥5
And ∵ 2x + 5 ≠ 0 (denominator cannot be equal to 0)
/ / X ≠ - 5 / 2
The value range is y ≤ 3 / 5
It can also be written as (- ∞, 3 / 5]
complete.

2 sin (2x + π / 6) let x ∈ (0, π / 4), and find the range of value of function f (x) x∈（0，π/4）， So 2x ∈ (0, π / 2) So 2x + π / 6 ∈ (π / 6,2 π / 3) So sin (2x + π / 6) ∈ (1 / 2,1] So 2Sin (2x + π / 6) ∈ (1,2] So the value range of F (x) is (1,2], so 2x + π / 6 ∈ (π / 6,2 π / 3) So sin (2x + π / 6) ∈ (1 / 2,1] Isn't sin2 π / 3 the radical 3 / 2? How about 1

A: because that function gets the maximum value at x = pi / 6, you can solve this problem as follows: y = (2x + π / 6) 2x + π / 6 ∈ (π / 6,2 π / 3), then sin y obviously gets the maximum value at y = π / 2. I hope it can help you, thank you

Given that x ∈ (0, π / 2), 2Sin ^ 2 (x) - sinxcosx-3cos ^ 2x = 0, find the value of sin (x + π / 4) / sin2x + cos2x + 1

∵2sin^2(x)-sinxcosx-3cos^2x=0
Divide both sides by cos? X at the same time
The result is: 2 Tan 2 x-tanx-3 = 0
ν TaNx = - 1 or TaNx = 3 / 2
∵x∈（0,π/2）,tanx>0
∴tanx=3/2
∴sinx/cosx=3/2
﹣ SiNx = 3 / 2 * cosx is replaced by sin? X + cos? X = 1
∴9/4*cos²x+cos²x=1
∴cos²x=4/13
∵x∈（0,π/2）∴cosx=2/√13
sin(x+π/4)/(sin2x+cos2x+1)
=(sinxcosπ/4+cosxsinπ/4)/(2sinxcosx+2cos²x)
=√2/2(sinx+cosx)/[2cosx(sinx+cosx)]
=√2/4*1/cosx
=√2/4*√13/2
=√26/8

Find y = sinxcosx + sin ^ 2x, 0

y=1/2sin2x-1/2cos2x=1/2
=1/2^1/2sin(2x-π/4)=1/2
When x = π / 2, y (max) = 1
When x = 0, y (min) = 0

Reduce the value of cos2x ^ 3x ^ 2x in the domain of cos2x ^ 2x

y=sin²x+2√3sinxcosx+3cos²x
y=sin²x+cos²x+2cos²x+√3sin(2x)
y=1+2cos²x+√3sin(2x)
y=2cos²x-1+2+√3sin(2x)
y=cos(2x)+√3sin(2x)+2
y=2[(1/2)cos(2x)+(√3/2)sin(2x)]+2
y=2[sin(π/6)cos(2x)+cos(π/6)sin(2x)]+2
y=2sin(π/6+2x)+2
y=2sin(2x+π/6)+2
The following values are obtained:
y=2sin(2x+π/6)+2
Because: - 1 ≤ sin (2x + π / 6) ≤ 1
Therefore: - 2 ≤ 2Sin (2x + π / 6) ≤ 2
Therefore, 0 ≤ 2Sin (2x + π / 6) + 2 ≤ 4
Therefore, the value range is: y ∈ [0,4]
In order to let the landlord see clearly, it is wordy

Simplify sin (- 2x)

sin(-2x)=-sin2x

Simplification of sin? X + 2sinx + cosx + 3cos? X

sin²x+2sinx*cosx+3cos²x
=1+2sinxcosx+2cos²x
=sin2x+cos2x+2
=√sin(2x+π/4)+2

What is the monotone interval of the function y = | x | and the symmetry axis of its image?

When x ≥ 0, y = x monotone increasing function
When x < 0, y = - x single adjustment and subtraction function
So (negative infinity, 0) monotone decreasing function, [0, positive infinity) monotone increasing function
The symmetry axis is x = 0, that is, the Y axis