Simplification of root sign half to root ten For example, if the root sign is half of the root sign two (please answer with a mathematical symbol), you can refer to the answer here http://zhidao.baidu.com/search?word= The first answer is to simplify the radical one to the root one hundred & TN = ikaslist & CT = 17 & ie = UTF-8 & SC = hao123 & RN = 20 It's like this: one-half of the root, one-third of the root, two-thirds of the root, one-quarter of the root, three-quarters of the root, one-fifth, two-fifths of the root Analogy, do you understand?

In other words, 1 / 3 of the root is equal to 3 / 3 of the root 3. 1 / 4 of the root is equal to 4 / 4 of the root 4

Given vector a = (2cosx, radical 3), B = (cosx, - SiNx) (1) When a is parallel to B, find the value of 2cos square x-sinx (2) Find the minimum and maximum of the function f (x) = a · B on [- π / 2,0]

(1) The vector a = (2cosx, Radix 3), B = (cosx, - SiNx) a ‖ B, so 2cosx / cosx = √ 3 / (- SiNx) i.e. SiNx = - √ 3 / 2, so 2cos? X-sinx = 2 (1-sin? X) - SiNx = 2 (1-3 / 4) - (- √ 3 / 2) = 1 / 2 + √ 3 / 2 (2) a * b = 2cos? X - √ 3sinx = 2 (1-sin? X) -

Given that a vector is equal to (cosx + SiNx, root 2cosx), B (cosx SiNx, Radix 2sinx), f (x) is equal to vector a × vector b (1) find the monotone interval of function f (x) (2) if 2x square - π x ≤ 0, find the value range of function f (x)

（1）
Vector a = (cosx + SiNx, √ 2cosx), B (cosx SiNx, √ 2sinx),
F (x) = vector a × vector b
=(cos²x-sin²x)+2sinxcosx
=sin2x+cos2x
=√2sin(2x+π/4)
From 2K π - π / 2 ≤ 2x + π / 4 ≤ 2K π + π / 2
K π - 3 π / 8 ≤ x ≤ K π + π / 8, K ∈ Z
The increasing range of F (x) is [K π - 3 π / 8, K π + π / 8], K ∈ Z
Similarly, the decreasing interval [K π + π / 8, K π + 5 π / 8], K ∈ Z is obtained
（2）
2 x 2 - π x ≤ 0 leads to 0 ≤ x ≤ π / 2
∴π/4≤2x+π/4≤3π/4
∴√2/2≤sin(2x+π/4)≤1
The value range of F (x) is [1, √ 2]

Given a = (root 2cosx, SiNx), vector b = (0, cosx), f (x) = | the square of vector a + vector B | 1. If 0

So sin2x = 1 / 2, x = 15 'cos (Ω - 15) = - cos30 = negative half of the root, a + B = (2cosx, SiNx) (0, cosx) = sinxcos x = 1 / 2, x = 1 / 2, x = 15' cos (Ω - 15) = - cos30 = the root number of the negative half, a + B = (root number 2cosx, SiNx + cosx), [a + b] = 2cosx ＾ 2 + (SiNx + cosx) ＾ 2 = 2cosx ＾ 2 + 1 + 2 sinxcos x = cos x = cos2x + sin2x-sin2x-sin2x-sin2x-sin2x-sin2x-sin2x-sin2x-sin2x-sin2x-sin2x1 = root 2Sin (2x + 45) - 1, so monotonically increasing interval - 3 / 8 Ω + K Ω

Given the vector a = (SiNx, cosx) B = (radical 3cosx, cosx) and B is not equal to 0, define the function f (x) = 2A · B-1 (1) Finding monotone increasing interval of function f (x) (2) If a is parallel to B, find the value of TaNx (3) If a is perpendicular to B, find the minimum positive value of X

f(x)=2a·b-1=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2sin(2x+π/6)
Monotone increasing interval of function f (x)
2x+π/6∈[(2k-1/2)π,(2k+1/2)π]
x∈[(k-1/3)π,(k+1/6)π]
If a is parallel to B, a × B = 0, √ 3cos? X-sinxcosx = 0, TaNx = √ 3, or does not exist
If a is perpendicular to B, a · B = 0, √ 3sinxcosx + cos? X = 0
2cosxsin(x+π/6)=0
X = (K + 1 / 2) π or (k-1 / 6) π
X minimum positive value
π/2

Given the vector a = (SiNx, cosx) vector b = (1, root 3), then | a + B | maximum value

A+b
= (= ( sinx.cosx )+(1,√3)
=(sinx+1,cosx+√3)
∴|a+b|=√[(sinx+1)^2+(cosx+√3)^2
=√[1+2sinx+1+2√3cosx+3]
=√[2(sinx+√3cosx)+4]
=√[4sin(x+60°)+5]
≤√(4+5)
=3
So the maximum value is 3

Given the vector M = radical 3sinx, cosx), P = (2 radical 3,1) if M is parallel to P, then SiNx * cosx=

If M is parallel to P, then:
Root sign 3sinx / 2 root sign 3 = cosx / 1
Then: SiNx = 2cosx
Because SiNx * SiNx + cosx * cosx = 1
Then: SiNx = 2 root sign 5 / 5, cosx = root 5 / 5
sinx*cosx=2/5

Let m = (cosx, SiNx), X ∈ (0, π), n = (1, radical 3) 1. If the modulus of M-N = radical 5, find the value of X 2. Let f (x) = (M + n) n, and find the value range of function f (x)

1.m-n=(cosx-1,sinx-√3）
|m-n|=√[（cosx-1）²+（sinx-√3）²]=√5
cos²x-2cosx+1+sin²x-2√3sinx+3=5
-2cosx-2√3sinx=0
TaNx = - √ 3 / 3, x = k π - π / 6
2.m+n=（cosx+1,sinx+√3）
f（x）=cosx+1+√3（sinx+√3）=2sin（x+π/6）+4
So the range is [2,6]

Let m = (cosx, SiNx), X belong to (0, PI), n = (1, 3 under the radical) (1) If | M-N | = 5 under the root sign, find the value of X (2) Let f (x) = (M + n) n, find the analytic formula of function f (x)

M-N = (cosx-1, SiNx radical 3)
|m-n|^2=5
That is (cosx-1) ^ 2 + (SiNx radical 3) ^ 2 = 5
Cos ^ 2x-2cosx + 1 + sin ^ 2x-2 radical sign 3sinx + 3 = 5
2cosx + 2 radical sign 3sinx = 0
1 / 2cosx + radical 3 / 2sinx = 0
sin(x+30)=0
Zero

Given the vector a (cosx, SiNx), B (Radix 2, Radix 2), ab = 8 If AB = 8 / 5, then cos (x-wu / 4) =?

Vector a dot multiplication vector b = 2 cos (x-wu / 4) = 8 / 5
Therefore, cos (x-wu / 4) = 4 / 5