It is proved that if the function f (x) satisfies the relation f '(x) = f (x) and f (0) = 1 in (- ∞, + ∞), then f (x) = e ^ X

It is proved that if the function f (x) satisfies the relation f '(x) = f (x) and f (0) = 1 in (- ∞, + ∞), then f (x) = e ^ X

Let f (x) = e ^ (- x) f (x), then f '(x) = 0, so f (x) is a constant, and f (0) = 1, so f (x) = 1, then f (x) = e ^ X

A problem of differential mean value theorem in higher numbers
F (x) is continuous in [0,1], differentiable in (0,1), and f (0) = f (1) = 0, f (0.5) = 1
It is proved that there is at least one m in (0,0.5) such that f (m) = M

Let f (x) = - 4 (x-0.5) ^ 2 + 1. Obviously, f (x) satisfies the problem condition. In (0,0.5), let g (x) = f (x) - x = - 4 (x-0.5) ^ 2 + 1-x = - 4x ^ 2 + 3x = - 4x (x-3 / 4) = x (0.75-x) > 0f (x) = - 4 (x-0.5) ^ 2 + 1 > x hold. Therefore, for this f (x), it does not exist in an m in (0,0.5), so that f (m) = M

Please tell me a math problem. How to calculate the number of minus 34 / 15 times 25

Minus 34 / 15 times 25 = - (34 * 25) / 15 = - 170 / 3

Let {an} be an equal ratio sequence with a common ratio greater than 1, if a1 + A2 + a3 = 7, and a1 + 3, 3a2, A3 + 4 constitute an equal ratio sequence, the general formula of the sequence {an} can be obtained

It seems that there is no real root, wrong question? A1 + A2 + a3 = 7, x09a2 = 7 - A1 - A3, x09a22 = A12 + A32 + 49 + a1a3-7a1-7a3a1xa3 = A22 = A12 + A32 + 49 + a1a3-7a1-7a3, x09a12 + a32-7a1-7a3 + 49 = 0 -------- ① (a1 + 3) (A3 + 4) = (3a2) 2 = 9a22 = 9a1a3, x098a1a3-4a1 = 3

Simple operation of 98 times 199

98×199
=98×（200-1）
=19600-98
=19502

As shown in the figure, if the median EF of the isosceles ladder ABCD is 6 and the median ad is 5, then the circumference of the isosceles ladder is ()
A. 11B. 16C. 17D. 22

∵ EF is the median line of ladder shaped ABCD, and EF = 6 ∵ AB + DC = 2ef = 2 × 6 = 12 ∵ the circumference of ladder shaped ABCD is ab + DC + AD + BD = 12 + 5 + 5 = 22, so D is selected

If you think about it, please join the following equations into equations by choosing the appropriate operation symbols or brackets
1 2 3 4=1 1 2 3 4 5=1 1 2 3 4 5 6=1
1 2 3 4 5 6 7=1 1 2 3 4 5 6 7 8=1

1*2+3-4=1
1*(2-3)*(4-5)=1
1+2*[(3-4)-(5-6)]=1
1+(2-3)*[(4-5)-(6-7)]=1
(1-2)*(3-4)*(5-6)*(7-8)=1

How to prove that all eigenvalues of symmetric matrix are real numbers

Let's talk about real matrix. Let's do it for comparison. A calls l eigenvalue E. eigenvector D denotes conjugate transpose (number ratio L, i.e. conjugate) AE = Le (1), then d (E) AE = LD (E) e = l | e | (2) (1) computes conjugate transpose D (E) a = D (L) d (E) then d (E) AE = D (L) d (E) AE = D (L) | e | (3) ratio (2) (3) (note both left and right numbers)

A three digit number is 19 times the sum of its digits. There are 11 such three digits. What is the difference between the smallest and the largest?

Suppose 100 digit is a, 10 digit is B and one digit is C
100A+10B+C＝19（A+B+C）
81A＝9B+18C
9A＝B+2C
A＝1 B＝1 C＝4
A＝1 B＝3 C＝3
A＝1 B＝5 C＝2
A＝1 B＝7 C＝1
A＝1 B＝9 C＝0
A＝2 B＝0 C＝9
A＝2 B＝2 C＝8
A＝2 B＝4 C＝7
A＝2 B＝6 C＝6
A＝2 B＝8 C＝5
A＝3 B＝9 C＝9
So the minimum is 114 and the maximum is 399

If a is a 4 * 3 matrix, the column vectors are linearly independent, and B is a third order invertible matrix, what is the rank of AB

If a is a 4 * 3 matrix and the column vectors are linearly independent, then the rank of matrix A is 3, that is, rank (a) = 3. B is a third order invertible matrix,
Multiplying by an invertible matrix does not change the rank, so rank (AB) = rank (a) = 3, that is, the rank of AB is 3