If the 2x power of 2 is multiplied by the 5x power of 64 = 2, then the x power of (x-3) =?

If the 2x power of 2 is multiplied by the 5x power of 64 = 2, then the x power of (x-3) =?

Because the 2x power of 2 times the 5x power of 64 = 2,
SO 2 ^ 2x * 2 ^ 6 = 2 ^ 5x,
2^（2x +6）=2^5x
2x+6=5x,
x=2,
So the x power of (x-3) = (2-3) ^ 2 = 1

If (X-2) 0 - (2x-6) - 3 is meaningful, then the range of X is () A. x＞2 B. x＜3 C. X ≠ 3 or X ≠ 2 D. X ≠ 3 and X ≠ 2

If the formula is meaningful, then x-3 ≠ 0 and 3x-6 ≠ 0,
The solution is: X ≠ 3 and X ≠ 2
Therefore, D

If the - 6th power of the root sign X-1 / 2x-3 + (X-2) is meaningful, then the value range of X

X-1 / 2x-3 ≥ 0, but 2x-3 ≠ 0
therefore
X ≤ 1 or X ＞ 3 / 2
x-2≠0
That is, X ≠ 2
therefore
The value range of X is:
X ≤ 1 or 3 / 2 ＜ x ＜ 2 or X ＞ 2

Cauchy mean value theorem proves that f (a) - f (m) / g (m) - G (b) = f '(m) / g' (m) f (x), G (x) satisfies the continuous differentiability in the interval a and B, and G '(x) is not equal to 0 M is the number in the interval {f (a) - f (m)} and {g (m) - G (b)} are in a bracket, which mainly means that the above is divided by the below.

prove:
Method 1
Do not remember that f (x) = g (x) [f (x) - f (a)],
Then f (x) and f (x) satisfy the Cauchy mean value theorem condition on [a, b],
It can be seen that at least one point m belongs to (a, b) so that
[F(b)-F(a)]/[f(b)-f(a)]=F'(m)/f'(m),
That is, G (b) = {G '(m) [f (m) - f (a)] + F' (m) g (m)} / F '(m), it is proved by sorting
Method 2
Note f (x) = [f (x) - f (a)] [g (x) - G (b)],
It is known that f (x) is continuous on [a, b], differentiable in (a, b), and f (a) = f (b) = 0. Therefore, f (x) satisfies Rolle's theorem on [a, b]. It can be seen that at least one point m belongs to (a, b) so that
F'(m)=0,
That is, f '(m) [g (m) - G (b)] + G' (m) [f (m) - f (a)] = 0, which is obtained by sorting

How to find the limit of (x's square-x) / SiNx by Cauchy mean value theorem

Is it when x tends to 0, if so
Let f (x) = x ^ 2-x,
G(x)=sinx,
F'(x) =2x-1,F'(0)=-1,
G'(x)=cosx,G'(0)=1,
When x tends to 0,
(x^-x)/sinx=(F(x)-F(0))/(G(x)-G(0))=F'(0)/G'(0)=-1

Proof of Cauchy mean value theorem When looking at the proof, I don't understand the auxiliary function value constructed when proving Lagrange's theorem. When I look at the two endpoints f (a) and f (b), I know that it is equal to 0 (because the connecting line at both ends intersects the original function) When proving Cauchy's mean value theorem, no image is given, but according to the condition, as long as the derivative of the second function is not 0, any other shape can be given, that is, there may be no intersection between the two functions in the interval [a, b], that is, the value of the two ends of the auxiliary function (the first function minus the second function) is not 0 However, the requirement for the application of Rawl's theorem is that the endpoint value is 0, which is not understood In the textbook, the function representing the value of the directed line segment Mn is used as the auxiliary function. The ordinate of point m is y = f (x), and the ordinate of point n is written as y = f (a) + [f (a) - f (b)] / (B-A) [f (x) - A]. In this way, the Lagrange median theorem is proved. In the book, it is proved that the ordinate of point n is y = f (a) + [f (b) - f (a)] / [f (b) - f (a)] / [f (x) - f (a)]. Why is it clear that point n is only in the second function f (x) How can it have something to do with F (x), not even the edge? How is this ordinate written

Do you still prove it with that? I don't know why the proof in the textbook is so poor. What can the pursuit of that geometric meaning help us understand? Just multiply and subtract diagonally,
order φ (x)=f(x)[g(b)-g(a)]-g(x)[f(b)-f(a)],
Easy to verify, φ （b）＝ φ (a) , just use Rolle's theorem