In triangle ABC, a (- 1,0) and C (1,0) are known. If a is greater than B and C, and 2sinb = Sina = sinc is satisfied Then the trajectory equation of vertex B

In triangle ABC, a (- 1,0) and C (1,0) are known. If a is greater than B and C, and 2sinb = Sina = sinc is satisfied Then the trajectory equation of vertex B

If Sina = sinc and a > C
Then a + C = 180 ° is in contradiction with the known one
You may have written it wrong, maybe 2sinb-sina = sinc
If so, then 2B = a + C is known from the sine theorem
If B = AC = 2, then a + C = 4
Let the coordinates of point B be (x, y)
Then root [(x + 1) ^ 2 + y ^ 2] + root [(x-1) ^ 2 + y ^ 2] = 4
The solution is x ^ 2 / 4 + y ^ 2 / 3 = 1
It's an ellipse
Given a = {x ﹤ a}, B = {1 ﹤ x ﹤ 2}, and Au (CRB) = R, then the value range of real number a is?
A.a≤1 B.a＜1 C.a≥2 D.a＞2
U is union
(CRB) is the complement of B
D
D
a>=2
In the known triangle ABC, a, B, C become A.P. prove: Sina + sinc = 2sinb
a. B, C is a. P, that is, 2b = a + C
According to the sine theorem, a / Sina = B / SINB = C / sinc = 2R is obtained
Therefore, 2 * 2rsinb = 2rsina + 2rsinc
Therefore, 2sinb = Sina + sinc
Let a = b-X, C = B + X
And Sina / a = SINB / b = sinc / C
sinA+sinC=sinB*(a+c)/b
=2sinB
If the root sign (3x-1) ^ 2 = 1-3x, then the value range of X is
3x-1≥0
3X≥1
X≥1/3
1-3x≥0
3X≤1
X≤1/3
So x = 1 / 3
Given △ ABC, a: B: C = 1:2:3, a = 1, then a − 2B + csina − 2sinb + sinc=______ .
According to a: B: C = 1:2:3, a = 30 °, B = 60 °, C = 90 °, ∵ a = 1, ∵ C = 2, B = 3, ∵ according to the sine theorem: Asina = bsinb = csinc = − 2B − 2sinb = 112 = 2, then a − 2B + csina − 2sinb + sinc = 2
It is known that the function f (x) = absolute value of loga (x + 1) (a > 0 and a ≠ 1) is monotonous if f (x) > 0 is constant when x belongs to (- 1,0)
Explain in what interval are increasing and decreasing functions,
It is known that the function f (x) = absolute value of loga (x + 1) (a > 0 and a ≠ 1) is monotonous if f (x) > 0 is constant when x belongs to (- 1,0)
Analysis: ∵ function f (x) = log (a, | x + 1 |) (a > 0 and a ≠ 1)
The domain of definition is x ≠ - 1
When x ∈ (- 1,0), f (x) > 0 holds
f(0)=log(a,|0+1|)=0
∴0
Given the quadratic function y = AX2 + BX + C, when x = 1, the maximum value of Y is 5, and its image passes through the point (2,3), find the relation of this function
Let the analytic formula of parabola be y = a (x-1) 2 + 5, substitute (2, 3) into a × (2-1) 2 + 5 = 3, and the solution is a = - 2, so the analytic formula of quadratic function is y = - 2 (x-1) 2 + 5 = - 2x2 + 4x + 3
Given the function f (x) = loga (x + 1), G (x) = loga (x-1) (a is greater than and not equal to 0), find the set of X with F (x) - G (2x) greater than 0
The original formula can be changed into
F (x) - G (2x) = loga [(x + 1) / (2x-1)] (a > 0, and a is not equal to 1)
If it is meaningful at this time, then (x + 1) / (2x-1) > 0, we can get: X1 / 2
Let's talk about a again
If 0 < a < 1
(x + 1) / (2x-1) < 1 satisfies the meaning of the problem to get x
If a > 1
(x + 1) / (2x-1) > 1 satisfies the meaning of the problem and gets X
F (x) - G (2x) = loga (x + 1) - loga (2x-1) = ln (x + 1) / lna-ln (2x-1) / LNA = [ln (x + 1) - ln (2x-1)] / LNA, because a is greater than or not equal to 0, then LNA > 0, f (x) - G (2x) > 0, ln (x + 1) - ln (2x-1) > 0, that is, x + 1 > 2x-1, x0,2x-1 > 0, x > 0.5, so the set of X {x | 0.5 that f (x) - G (2x) is greater than 0 holds
Seeking the range of several formulas about quadratic function y = AX2 + BX + C
2a-b, 9a-4b, (ABC are all less than 0). 2c-3b, a + B-M (am + b) (M is not equal to 1) (a is less than 0, BC is greater than 0) the relationship between these formulas and 0 (greater than or equal to 0)
It's all written, and it's an additional 30 points
2a-b＜0
9a-4b＜0
2c-3b＞0
a+b-m(am+b)＞0
Why don't you understand
What do you mean
Given the function f (x) = loga (x + 1), G (x) = loga (4-2x) (a > 0, and a ≠ 1); (I) find the domain of definition of function y = f (x) - G (x); (II) find the range of values of X that make the value of function y = f (x) - G (x) positive
(I) we can get x + 1 ＞ 04 − 2x ＞ 0 and the solution is - 1 ＜ x ＜ 2. We can get that the definition domain of function f (x) is (- 1,2). (II) f (x) = f (x) - G (x) = log a (x + 1) - log a (4-2x) = log a & nbsp; X + 14 − 2x, & nbsp; When a ＞ 1, from x + 14 − 2x ＞ 1 − 1 ＜ x ＜ 2, the solution is 1 ＜ x ＜ 2, so the value range of X is (1,2). When 0 ＜ a ＜ 1, from 0 ＜ x + 14 − 2x ＜ 1 − 1 ＜ x ＜ 2, the solution is - 1 ＜ x ＜ 1, so the value range of X is (- 1,1)