X1 is the root of the equation: X * lgx = 2006; X2 is the root of the equation: X * (10 ^ x) [x power of 10] = 2006; find the value of X1 * x2

X1 is the root of the equation: X * lgx = 2006; X2 is the root of the equation: X * (10 ^ x) [x power of 10] = 2006; find the value of X1 * x2

This problem is a transcendental equation. We can't solve X1 and X2 directly. We can only find out the law
From the meaning of the title, we can know: X1 * lgx1 = 2006 ············································································,
x2*(10^x2)=2006··················（2）；
Comparing equation (1) (2), we find that X1 = 10 ^ x2 ··············· (3),
Equation (2) can be written as: 10 ^ x2 = 2006 / (x2) ·························································································;
So from (3) (4), we can get: X1 = 10 ^ x2 = 2006 / (x2),
Namely: X1 * x2 = 2006

Let the root of equation 10 ^ x + x-3 = 0 be α, and the root of equation lgx + x-3 = 0 be β, then the value of α + β is?

The root of 10 ^ x + x-3 = 0 is α, so the root of 3 = 10 ^ α + α ----- (1) lgx + x-3 = 0 is β, so 3 = LG β + β ----- (2) from (1) (2), we get: LG β + β = 10 ^ α + α, that is: LG β = 10 ^ α + α - β, that is: β = 10 ^ (10 ^ α + α - β), that is: β = 10 ^ (10 ^ α + α) / (10 ^ β), that is: 10 ^ (10 ^ α) * 10 ^ α = (...)

The interval where the equation X-1 = lgx must have a root is ()
A. （0.1，0.2）B. （0.2，0.3）C. （0.3，0.4）D. （0.4，0.5）

Let f (x) = x-1-lgx, then f (0.1) = 0.1-1-lg0.1 = 0.1.1, f (0.2) = 0.2-1-lg0.2 = 0.2-1-1-lg0.2 = 0.2-1 - (lg2-1-1-lg2-1-1-lg0.2 = 0.2-1-lg0.2, f (0.1) = 0.1-1-1-lg0.1-lg0.1.1.1-lg0.1, then f (0.1) = 0.1-1-1-lg0.1-1-lg0.1-1-1-lg0.1-1-1-1-1-lg0.2-1-1-1-lg0.2, f (0.2) (0.2, f (0.2, f (0.1) (0.1) (0.2) (0.2) (0.2) (0.3-3 1

The relationship between the radius of circumcircle and three sides of right triangle
The three sides are 9, 40 and 41 respectively

- -|||
Because 9 & sup2; + 40 & sup2; = 41 & sup2;
So a triangle with sides 9, 40 and 41 is a right triangle
41 is the side opposite the right angle
So 41 is the diameter of the circumscribed circle
That is radius r = 20.5

Let u = x - 2 ≤ x ≤ 4 and a = x 0

[reference answer]
A intersection B = {XL 1 ≤ x

As shown in the figure, in △ ABC, ∠ BAC = 108 ゜, ab = AC, BD bisects ∠ ABC, intersects AC with D, and proves: BC = CD + ab

In △ abd and △ EBD, in △ abd and △ EBD, in △ abd and △ EBD, in △ abd and △ EBD, ab = EB, ab = EB, abd = abd = ebdbd = BD, be = Ba = Ba, connect De, De, \8780; ≌ EBD (SAS), connect de, and W BD 87; BD, bisbisbisbisbisbisbisbisbisbisbisbisbisbisbisdivide \\\\A BD