What is the absolute value of 3 minus root 10?

What is the absolute value of 3 minus root 10?

Root ten minus three

Find the indefinite integral of (x ^ 3 + 2x ^ 2 + 1) / (x-1) (X-2) (x-3) ^ 2

0

0

Let x = sin T, the original formula = ∫ (0, π / 2) sin T ^ 2cost ^ 2DT = ∫ (0, π / 2) (sin2t) ^ 2 / 4dt = ∫ (0, π / 2) 1 / 8 (1-cos4t) = π / 16

Let f (x) = cos ω X（ 3 sin ω x + cos ω x), where 0 ＜ ω ＜ 2 (I) if the period of F (x) is π, if - π 6≤x≤π When 3, find the range of F (x); (II) if one of the symmetry axes of the graph of function f (x) is x = π 3. Find the value of ω

（Ⅰ）f(x)＝
3sin ωxcosωx+cos2ωx=sin(2ωx+π
6)+1
Two
∵T=π，ω＞0
∴2π
2ω＝π
∴ω=1
When − π
6≤x≤π
3 is 2x + π
6∈[ −π
6，5π
6] Sin (2x + π)
6)∈[−1
2，1]
∴f(x)∈[0，3
2]
The value range of F (x) is [0,3]
2]
（Ⅱ）f(x)＝sin(2ωx+π
6)+1
The axis of symmetry of 2 is x = π
Three
∴2ω×π
3+π
6＝kπ+π
2，K∈Z
∴ω＝3K+1
Two
∵0＜ω＜2
∴−1
3＜K＜1      
k=0，ω＝1
Two

Find the minimum value of the function f (radical x) = X-1

F (radical x) = X-1
f(x)=x ^ 2 - 1
When x = 0, take the minimum value
f(x)min = -1

The upper limit LN2 and the lower limit 0 (root e ^ x-1) DX of definite integral ∫ are obtained,

Let √ (e ^ x-1) = t, then DX = 2tdt / (1 + T?)
∵ when x = LN2, t = 1. When x = 0, t = 0
The original formula = 2 ∫ (0,1) t? DT / (1 + T?)
=2∫(0,1)(1-1/(1+t²))dt
=2(t-arctant)|(0,1)
=2(1-π/4)
=2-π/2

10. Where y is the real number root sign (1 + x) + root sign (1-y) = 0, find the 2012 power of X + the 2012 power of Y

The sum of two nonnegative terms is 0, and both nonnegative terms are equal to 0
1+x=0 x=-1
1-y=0 y=1
x^2012+y^2012
=(-1)^2012+1^2012
=1+1
=2

If x and y are real numbers, and the absolute value of X + 2 = Y-2 under the negative root sign, then what is x divided by y to the power of 2012

The absolute value of X + 2 = Y-2 under negative radical
Under the absolute value of X + 2 + radical, Y-2 = 0
x+2=0
x=-2
y-2=0
Y=2
(x÷y)
=The 2012 power of (- 2 △ 2)
=The 2012 power of (- 1)
=1

x. If y is a real number, and y = 2x + 4 + 4 + 4-2x + 3, then find the y-th power of X

Because y = √ (2X-4) + √ (4-2x) + 3, and X, y are real numbers, so 2X-4 ≥ 0, 4-2x ≥ 0, the solution is x = 2, y = 3, so x ^ y = 8

Mathematical problem (given that a is an integer less than 4 + Radix 2, and the absolute value of 2 minus a is equal to a minus 2, find all possible values of A Given that a is an integer less than 4 + radical 3, and the absolute value of 1 minus a is equal to a minus 1, find all possible values of A Excuse me~

Because the absolute value of 1 minus a is equal to a minus 1, A-1 < or = 0, so a > or = 1
A is an integer less than 4 + root 3, so a less than 5.732 is 1,2,3,4,5