It is known that ellipse C1 and hyperbola C2 have the same focus F1 (- 3,0) and F2 (3,0), and both pass through D (2, radical 2) Let a (x, y) be the point on the ellipse C1. If the minimum distance between a and B (x0,0) (x0 > 0) is not less than 2 root sign 3 / 3, the value range of x0 is obtained

It is known that ellipse C1 and hyperbola C2 have the same focus F1 (- 3,0) and F2 (3,0), and both pass through D (2, radical 2) Let a (x, y) be the point on the ellipse C1. If the minimum distance between a and B (x0,0) (x0 > 0) is not less than 2 root sign 3 / 3, the value range of x0 is obtained

This is an ellipse, which is the ellipse x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\= B & # 178; + 9, that is: B =
It is known that the positive scale function y is equal to KX, when x increases by 3, y decreases by 2, and the value of K is obtained
y = kx
y-2 = k(x+3) ,y= kx + 3k +2
So, 3K + 2 = 0, k = - 2 / 3
If the logarithm of log with base 3 T = 2 + sin α, then | T-1 | + | T-9 | =?
Such as the title
1≤2+sinα≤3,
So 3 ≤ t ≤ 9, so T-1 > 0, T-9 ≤ 0,
So | T-1 | + | T-9 | = T-1 + 9-t = 9
When x increases by 3, y decreases by 4, and the value of K is obtained
y-4=k(x+3) （1）
y=kx （2）
Replace (2) with (1)
kx-4=kx+3k
k=-4/3
-4/3
Do with special value
Let x = 1 then y = K
Then when x = 4, y = 4K = K-4
The solution is k = - 4 / 3
So y = - 4 / 3 X
test
When x '= x + 3
y'=-4/3(x+3)=-4/3x -4 =y-4
So k = - 4 / 3 holds
Let two points, (x1, Y1) (X2, Y2) be suitable for this function, so Y1 = kx1, y2 = kx2. If you want to subtract the two formulas, you can get y2-y1 = K (x2-x1), so k = (y2-y1) / (x2-x1) = - 4 / 3
So k = - 4 / 3
Find the logarithm of cos π / 9 with log base 2 + log base 2 Cos2 π / 9 + log base 2 4 π / 9=
If the logarithm of the same base is added, the result is equal to the logarithm of the product of true numbers
The product of real numbers is
cosπ/9cos2π/9cos4π/9
=(8sinπ/9cosπ/9cos2π/9cos4π/9)/(8sinπ/9)
=4sin2π/9cos2π/9cos4π/9)/(8sinπ/9)
=(2sin4π/9cos4π/9)/(8sinπ/9)
=(sin8π/9)/(8sinπ/9)
=(sin(π-π/9))/(8sinπ/9)
=1/8
So log2 (1 / 8) = - 3
When x = - 6, y = 3, then the analytic expression of the positive proportion function is?
When x = - 6, y = 3 is substituted into the function y = KX,
3=-6k
We can get k = - 1 / 2
So the analytic expression of the positive proportion function is y = - X / 2
Given that the logarithm of log with a as the bottom 4 / 5 is less than 1, find the value range of A
When a > 1, loga4 / 54 / 5, so a > 1
When 0
A
It is known that y is a positive proportional function of X. when the independent variable x increases by 2, the corresponding value of Y increases by 4, and the value of scale coefficient K is calculated
Because y = KX
So y + 4 = K (x + 2)
y+4=kx+2k
Because KX = y
So 4 = 2K
K=2
Let y = KX
y+4=k（x+2）
y+4=kx+2k
4=2k
K = 2. What if x is reduced by 2
If 0 ≤ log, the logarithm of X with base 2 < x; x < 2, then the value range of X is
solution
Because log (2) 1 = 0, log (2) 2 ^ 2 = log (2) 4 = 2
And log (2) x is an increasing function and 0 ≤ log (2) x < 2
So the value range of X is 1 ≤ x < 4
It is known that y is a positive proportion function of X, and when x = 2, y = 12. Find the proportion coefficient between Y and X, and write the function analytic formula of Y guanyintu X
Because y is a positive proportional function of X
therefore
y=kx
Because when x = 2, y = 12
therefore
12=2k
K=6
therefore
y=6x
The scale factor is 6
Since y and X are in positive proportion, let the analytic expression of the positive proportion function be y = KX (K ≠ 0),
When x = 2, y = 12, k = 6,
Therefore, the analytic formula of positive proportion function is y = 6x