Sin a = (A-3) / (a + 5), cos a = (4-2a) / (a + 5), a ∈ (π / 2, π) then the value of real number a

Sin a = (A-3) / (a + 5), cos a = (4-2a) / (a + 5), a ∈ (π / 2, π) then the value of real number a

(sina)^2+(cosa)^2=1
(a-3)^2/(a+5)^2+(4-2a)^2/(a+5)^2=1
(a-3)^2+(4-2a)^2=(a+5)^2
a^2-6a+9+4a^2-16a+16=a^2+10a+25
4a^2-32a=0
a=0,a=8
a∈（π/2,π）
So Sina > 0, cosa

Given the function f (x) (x is not equal to 0), for any nonzero real number x, y, f (XY) = f (x) + F (y)
(1) Find the value of F (1) and f (- 1);
(2) Judge the parity of y = f (x);
(3) If y = f (x) is an increasing function on (0, positive infinity) and f (x) + F (1-1 / x) is less than or equal to 0, the value range of X is obtained

Let's take F (XY) = f (x) + F (y) (1) let x = y = 1, then f (1x1) = f (1) + F (1) = 2F (1) so f (1) = 0, let x = y = - 1, then f (1) = 2F (- 1) f (- 1) = 0 (2) let y = - x, then f (- x) = f (x) + F (- 1) = f (x) odd and even, let's not mention it. (3) f (x) + F (1-1 / x) ≤ 0, so f (x

It is known that the focus of parabola C & # 8321; coincides with the right focus of ellipse C2: X & # 178 / 6 + Y & # 178 / 5 = 1, and the vertex of parabola C & # 8321; is in the original coordinate
Point, straight line L passing through point m (4,0) and parabola C & intersect with two points a and B respectively. If | ab | = 10 under 4 times root sign, the equation of straight line l can be solved

If the right focus of the ellipse is F2 (1,0), then the focus coordinate of the parabola is (1,0), then p / 2 = 1, P = 2, that is, the parabola equation is y ^ 2 = 4x. Let a coordinate be (x1, Y1), B (X2, Y2), and the equation of the straight line l be y = K (x-4). Substituting into the parabola equation is: K ^ 2 (x ^ 2-8x + 16) = 4xk ^ 2x ^ 2 - (8K ^ 2 + 4) x + 16K ^ 2 = 0x1 +

Let the ellipse pass through a fixed point, then the minimum distance between the center of the ellipse and the guide line

The directrix is x = B / A. the limit method is used in drawing. If a and B are almost equal, a circle will be formed. The length is equal to the radius of the circle and the distance from the origin to the fixed point

X 1 and x 2 are the two real roots of the equation x & # 178; - (2m-1) x + (M & # 178; + 2m-4) = 0. Find the minimum value of the square of x 1 + the square of x 2

X1 and X2 are the two real roots of the equation x & # 178; - (2m-1) x + (M & # 178; + 2m-4) = 0,
According to Weida's theorem, there are: X1 + x2 = 2m-1, X1 * x2 = M & # 178; + 2m-4
And △ = [- (2m-1)] and#178; - 4 * 1 * (m and#178; + 2m-4) > = 0, the solution is m

Let X1 and X2 be the two real roots of the equation x square - 2mx + (m square + 2m + 3) = 0, then the minimum value of X1 square + x2 square

According to Weida's theorem, because a = 1, B = - 2m, C = m ^ 2 + 2m + 3, so X1 + x2 = 2m, X1 x2 = m ^ 2 + 2m + 3, so X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1 x2 = 2m ^ 2-4m-6, by △ = B ^ 2-4ac = 4m ^ 2-4 (m ^ 2 + 2m + 3) = - 8m-12, and because this equation has two real roots and is the minimum, so - 8m-12 = 0, the solution is m = - 1

When m is greater than 0, find the minimum value of the square of (M + 1) X-2 (M's Square-1) x + m + 2m

y=(m+1)x²-2(m²-1)x+m²+2m
m> M + 1 > 1 at 0
Obviously, the opening is upward and there is a minimum value, which is obtained at the axis of symmetry
The axis of symmetry is x = 2 (M & # 178; - 1) / 2 (M + 1) = M-1
So the minimum value is y (m-1) = (M + 1) (m-1) &# 178; - 2 (M & # 178; - 1) (m-1) + M & # 178; + 2m = - M & # 179; + 2m & # 178; + 3m-1
If you don't understand, please hi me, I wish you a happy study!

x. Y is the two real roots of the equation m2-2am + A + 6 = 0, then the minimum value of (x-1) ^ 2 + (Y-1) ^ 2 is ()
A49/4 B-49/4 C8 D-6

If a = 3 / 4, the equation has no real root
Choose C 8
From the relationship between root and coefficient
x+y=2a
xy=a+6
Expand (x-1) ^ 2 + (Y-1) ^ 2
=x^2+y^2-2(x+y)+2
=(x+y)^2-2xy-2(x+y)+2
=4a^2-6a-10
Since the equation has two real roots, the discriminant △ = 4A ^ 2-4 (a + 6) ≥ 0 is expanded
△=4a^2-4(a+6)
=4a^2-4a-24
=4(a^2-a-6)
=4(a-3)(a+2)≥0
The solution of this inequality is: a ≥ 3 or a ≤ - 2,
Under the restriction of a ≥ 3 or a ≤ - 2, the minimum value of the following formula can be obtained:
(x-1)^2+(y-1)^2
=4a^2-6a-10
=2(2a^2-3a-5)
=2(2a-5)(a+1)
Consider the parabola with F (a) = 4A ^ 2-6a-10. The opening is upward, and the two intersections with the horizontal axis are: (- 1,0) (5 / 2,0). Under the limited conditions, its effective region is a ≥ 3 or a ≤ - 2, so the minimum value should be obtained at a = 3 or a = - 2,
The results are as follows
When a = 3, (x-1) ^ 2 + (Y-1) ^ 2 = 4A ^ 2-6a-10 = 36-18-10 = 8;
When a = - 2, (x-1) ^ 2 + (Y-1) ^ 2 = 4A ^ 2-6a-10 = 16 + 12-10 = 18;
Obviously, when a = 3, the minimum value of (x-1) ^ 2 + (Y-1) ^ 2 is: 8

Let X and X be two of the equations m2-2am + A + 6 = 0 about M, then the minimum value of (x-1) 2 + (Y-1) 2 is

There is something wrong with the title!
Let X and y be the two real roots of the equation m2-2am + A + 6 = 0, then the minimum value of (x-1) ^ 2 + (Y-1) ^ 2
From the relationship between root and coefficient
x+y=2a
xy=a+6
Expand (x-1) ^ 2 + (Y-1) ^ 2
=x^2+y^2-2(x+y)+2
=(x+y)^2-2xy-2(x+y)+2
=4a^2-6a-10
Since the equation has two real roots, the discriminant △ = 4A ^ 2-4 (a + 6) ≥ 0 is expanded
△=4a^2-4(a+6)
=4a^2-4a-24
=4(a^2-a-6)
=4(a-3)(a+2)≥0
The solution of this inequality is: a ≥ 3 or a ≤ - 2,
Under the restriction of a ≥ 3 or a ≤ - 2, the minimum value of the following formula can be obtained:
(x-1)^2+(y-1)^2
=4a^2-6a-10
=2(2a^2-3a-5)
=2(2a-5)(a+1)
Consider the parabola with F (a) = 4A ^ 2-6a-10. The opening is upward, and the two intersections with the horizontal axis are: (- 1,0) (5 / 2,0). Under the limited conditions, its effective region is a ≥ 3 or a ≤ - 2, so the minimum value should be obtained at a = 3 or a = - 2,
The results are as follows
When a = 3, (x-1) ^ 2 + (Y-1) ^ 2 = 4A ^ 2-6a-10 = 36-18-10 = 8;
When a = - 2, (x-1) ^ 2 + (Y-1) ^ 2 = 4A ^ 2-6a-10 = 16 + 12-10 = 18

Let X and y be the two real roots of the equation m2-2am + A + 6 = 0, then the minimum value of (x-1) 2 + (Y-1) 2 is ()
A. -1214B. 18C. 8D. 34

Then (x-1) 2 + (Y-1) 2 = x2 + y2-2 (x + y) + 2 = (x + y) 2-2xy-2 (x + y) + 2 = (2a) 2-2 (a + 6) - 4A + 2 = 4a2-6a-10 = 4 (a-34) 2-494