Let θ∈ R, 0

Let θ∈ R, 0

The solution set of cos θ x ^ 2 + 2xsin φ (sin θ + cos θ) + sin θ > 0 is (1,10)
It shows that the opening of quadratic function is downward, and the function passes through (1,0) and (10,0)
therefore
cosθ

Given the vector a = (sin95, cos95), B = (sin25, - sin65), then a * B is online

a*b=sin95*sin25-cos95*sin65
=sin95*sin25-cos95*sin(90-25)
=sin95*sin25-cos95*cos25
=-cos(95+25)
=-cos120
=0.5
It's detailed enough,

Let A. B be a real number and satisfy a = (√ B-3) + (√ 3-B) + 2. Find the value of √ ab · √ AB + 1 / A + B

a=√(b-3) +√(3-b)+2
Because √ (B-3) > = 0, √ (3-B) > = 0
That is, b > = 3, B

After factoring the following polynomials, the results containing the same factor are 16x ^ 5-x; (x-1) ^ 2-4 (x-1) + 4; (x + 1) ^ 4-4x (x + 1) ^ 2 + 4x;
-4X ^ 2-1 + 4x; what does it mean to have the same factor? Should the letter coefficient symbols be the same? Don't forget to explain after you finish,

（1）16X^5-X
=X(16X^4-1)
=X(4X^2+1)(4X^2-1)
=X(4X^2+1)(2X+1)(2X-1)
(2)(X-1)^2-4(X-1)+4
=X^2-2X+1-4X+4+4
=X^2-6X+9
=(X-3)^2
(3)=（X+1)^2 [(X+1)^2-4X]+4X
=(X+1)^2(X-1)^2+4X
-4X^2-1+4X=-[(2X^2)-4X+1]=-(2X-1)^2
With the same factor, e.g
a^2-4=(a+2)(a-2)
a^2+a-6=(a+3)(a-2)
A-178; - 4 is the same as a-178; + a-6, and the factor is A-2
It's hard to explain. Please forgive me if there are mistakes