If the power of (2013) - 2 is the power of (- 2013 +)

If the power of (2013) - 2 is the power of (- 2013 +)

（-2)^2013+(-2)^2014
=x+(-2)^2013*(-2)
=x-2x
=-x=2^2013

What is the third power of (the cube root of minus 12),

This formula is 3
So the answer is - 12

If a+ A2-2a + 1 = 1, then the value range of a is___ .

∵a+
a2-2a+1=1，
Qi
(a-1)2=-（a-1），
∴|a-1|=-（a-1），
∴a-1≤0，
∴a≤1．
So the answer is a ≤ 1

-The 2006 power of 2 plus the 2007 power of - 2? -What is the 2006 power of 2 plus the 2007 power of - 2?

(-2)^2006+(-2)^2007
=(-2)^2006+(-2)^2006*(-2)
=(-2)^2006*(1-2)
=-2^2006

If the third root sign 3x-7 and the third root sign 3Y + 4 are opposite numbers to each other, find the value of the third root sign x + y

The third root sign 3x-7 and the third root sign 3Y + 4 are opposite numbers
Then 3x-7 and 3Y + 4 are opposite numbers
So the sum is zero
3X-7+3Y+4=0
3X+3Y=3
X+Y=1
So the third root x + y = 1

Given that y = radical X-2 + Radix 2-x + 2, find the arithmetic square root of X of Y

x-2>=0,2-x>=0
∴x=2,y=2
Y to the power of x = 4
The arithmetic square root is 2

Given the absolute value of x = root 15, find the real number X

Remember: | a | = ± a
|x|=√15
x=±√15
Answer: x = ± √ 15

If the function y = sin ω x (ω > 0) has at least 10 maxima in the interval [0, 1], then the minimum value of ω is______ .

According to the image characteristics of sine function, there are 10 maximum values of the function, at least 91
4 cycles
There are at least 10 maxima in the interval [0,1] by the mean number y = sin ω x (ω > 0)
Then 91
4T≤1⇒37
4•2π
ω≤1，
We can get ω ≥ 37 π
Two
So the answer is: 37 π
2．

Find the maximum and minimum of the function y = 2cos? X + 2sinx-3

Let SiNx = 2 cos 2 x + 2 sinx-3 = 2 (1-sin ^ 2x) + 2sinx-3 = - 2Sin ^ 2x + 2sinx-1
Y = - 2T ^ 2 + 2t-1, t ∈ [- 1,1]. Symmetry axis t = 1 / 2
So the maximum value is y (1 / 2) = - 1 / 2. The minimum value is: min (Y (- 1), y (1)) = min (- 5, - 1) = - 5
Therefore, the maximum of the function is: - 1 / 2, and the minimum is: - 5

The fourth power of sin α - Sin 2 α + cos 2 α is expressed by cos α 2 (1-sin α) (1 + cos α) = (1-sin α + cos α) 2 It was proved that sin? α + sin? β - sin? α sin? β + cos? α cos? β = 1

sin^4α-sin^2α+cos^2α
=sin^2α(sin^2α-1)+cos^2α
=-sin^2αcos^2α+cos^2α
=cos^2α(1-sin^2α)
=cos^4α
2(1-sinα)(1+cosα)
=2(1+cosα-sinα-sinαcosα)
=2+2cosα-2sinα-2sinαcosα
=(sinα-cosα)^2-2(sinα-cosα)+1
=(sinα-cosα-1)^2
=(1-sinα+cosα)^2
sin^2α+sin^2β-sin^αsin^2β+cos^2αcos^2β
=sin^2α(1-sin^2β)+sin^2β+cos^2αcos^2β
=sin^2αcos^2β+sin^2β+cos^2αcos^2β
=cos^2β(sin^2α+cos^2α)+sin^2β
=cos^2β+sin^2β
=1