Let f (x) = loga (1-x) g (x) = loga (1 + x)! Let f (x) = loga (1-x) g (x) = loga (1 + x), (a > 0, a ≠ 1) The number of real roots of the equation a ^ (g (x-x ^ 2 + 1)) = a (f (k)) - x about X is discussed The number of real roots of the equation a ^ (g (x-x ^ 2 + 1)) = a ^ (f (k)) - x about X is discussed

Let f (x) = loga (1-x) g (x) = loga (1 + x)! Let f (x) = loga (1-x) g (x) = loga (1 + x), (a > 0, a ≠ 1) The number of real roots of the equation a ^ (g (x-x ^ 2 + 1)) = a (f (k)) - x about X is discussed The number of real roots of the equation a ^ (g (x-x ^ 2 + 1)) = a ^ (f (k)) - x about X is discussed

What is k
A (f (k))? Is it a ^ (f (k))
If the title is
Let f (x) = loga (1-x) g (x) = loga (1 + x), (a > 0, a ≠ 1)
This paper discusses the equation a ^ (g (x-x ^ 2 + 1)) = a ^ (f (k)) - x about X when the number of real roots K is r
Then:
It is concluded that X & # 178; - 2x-1-k = 0 has a solution in R, and K ＞ - 2 is obtained by using the discriminant
Two different solutions when k ＞ - 2
A solution of k = - 2
There is no solution for K ＜ - 2

If the logarithmic function f (x) satisfies f (2x) = 1 + F (x), then f (x)=

Undetermined coefficient method
Let f (x) = loga (x)
Substituting f (2x) = 1 + F (x)
∴ loga(2x)=l+loga(x)=loga(ax)
∴ a=2
That is, f (x) = log2 (x)

The sum of four consecutive natural numbers is less than 24. How many groups of such natural numbers are there?

There were five groups: 0123, 1234 / 2345 / 3456 / 4567
Let the first number be x, then the other three numbers are x + 1, x + 2, x + 3,
From the meaning of the title, we can get: x + (x + 1) + (x + 2) + (x + 3)

The sum of the reciprocal of four natural numbers is 1, and the four natural numbers are different from each other

1/2+1/4+1/6+1/12=1;
1/2+1/3+1/9+1/18=1;
1/2+1/3+1/10+1/15=1;
1/2+1/3+1/8+1/24=1；

The sum of three consecutive natural numbers is less than 12. Such a natural array has () groups
A. 1B. 2C. 3D. 4

Let the three continuous natural numbers be n, N + 1, N + 2. N + N + 1 + N + 2 ＜ 12n ＜ 3, so the three continuous natural numbers can be 0, 1, 2 or 1, 2, 3 or 2, 3, 4

The sum of three consecutive natural numbers is not more than 12. How many groups of such natural numbers are there?

The sum of three consecutive natural numbers is less than 12. How many groups are there
Three groups. (0,1,2), (1,2,3), (2,3,4)

The sum of three consecutive natural numbers is greater than 12 and less than 24. How many groups of such natural numbers are there?

(24-12) / 3-1 = 3 (Group) I'm sure there are three groups, namely 4, 5, 6; 5, 6, 7; 6, 7, 8

The sum of three consecutive natural numbers is less than 120. How many groups of such natural numbers are there? Why?
Why? Be sure to explain why!!!!! Don't just have answers!

38 groups
Let three continuous natural numbers be n-1, N, N + 1, the sum of which is (n-1) + N + (n + 1) = 3N

The sum of three consecutive natural numbers is less than 15. How many groups of such natural numbers are there?
Three or four? 0,1,

012 yes, isn't 0 a natural number now