If f (x) = a ^ 2-3x (a > 0, and a ≠ 1), G (x) = a ^ X 1. Calculate the fixed-point coordinates of the image of the function f (x); 2. Prove {g (x1 + x2)} / 2 ≤ {g (x1) + G (x2)} / 2

If f (x) = a ^ 2-3x (a > 0, and a ≠ 1), G (x) = a ^ X 1. Calculate the fixed-point coordinates of the image of the function f (x); 2. Prove {g (x1 + x2)} / 2 ≤ {g (x1) + G (x2)} / 2

1. In order to make (2-3x) times of a a a fixed value, we must make (2-3x) = 0, that is, x = 2 / 3
Take x = 2 / 3 into the function and get f (2 / 3) = 1
So constant crossing point (2 / 3,1)
2. If we take g (x) = a ^ x, we get
a^{(x1+x2)/2}≤(a^x1+a^x2)/2
Reduced 2 ≤ a ^ {(x1-x2) / 2} + A ^ {(x2-x1) / 2}
=a^{（x1-x2）/2}+(1/a)^{(x1-x2)/2}
Just prove this, according to a + B ≥ 2 radical (AB), you can easily get the answer

Given the function f (x) = x + 1 / X (x > 0), the minimum distance between the point on the image of function f (x) and the line 3x-4y-1 = 0 is obtained, and the coordinates of the corresponding point are obtained

D = | 3x-4 (x + 1 / x) - 1 | / 5 = 1 / 5 * | x + 4 / x + 1 | = x + 4 / x + 1; mean inequality
The minimum value is Dmin = 2 * 2 + 1 = 5; when x = 4 / x, i.e. x = 2, the corresponding points are (2, f (2)), (2,2.5)

Simple calculation of 48 × 15 △ 25-16

48×15÷（25-16）
=48×15÷9
=48×5/3
=80

Let the range of function y = 4x-2x + 1 + 2 be d when x ≤ 1, and f (x) = x2 + KX + 5 ≤ 4x when x ∈ D

Let t = 2x, because x ≤ 1, then t ∈ (0, 2], then the original function y = t2-2t + 2 = (t-1) 2 + 1 ∈ [1, 2], that is, d = [1, 2] from the topic meaning: F (x) = x2 + KX + 5 ≤ 4x, method 1: then X2 (K-4) x + 5 ≤ 0, when x ∈ D, it always holds ﹣ 1 + (K − 4) + 5 ≤ 022 + (K − 4) 2 + 5 ≤ 0 ﹣ K ≤ − 2K 12 ﹣ K ≤ - 2

This rectangular biscuit box is 17cm in length, 10cm in width and 20cm in height. If a circle of brand paper is pasted on the side of the box, what is the area of the brand paper at least
Rice?
This rectangular biscuit box is 17cm in length, 10cm in width and 20cm in height. If a circle of brand paper is pasted on its side, how many square centimeters should the brand paper cover?

(17 + 10) * 2 * 2 = 108 square centimeter
If there is any help, please take it. Thank you
This is the first answer. I wish you progress!

What are the four operation symbols to be filled in the middle of 5? At last, they are equal to 24

5*5-5/5=25-1=24;

Let the sum of the first n terms of the equal ratio sequence {an} be Sn, and S1 = 18, S2 = 24, then S4 equals ()
A. 763B. 793C. 803D. 823

If q = 1, then S2 = 2S1, obviously 24 = 2 × 18 does not hold, so Q ≠ 1. From S1 = 18, S2 = 24, A1 = 18, a1 + A2 = 24, so A2 = 6, so common ratio q = a2a1 = 618 = 13. So S4 = A1 (1 − Q4) 1 − q = 18 × (1 − (13) 4) 1 − 13 = 803

Use a 15 cm long, 4 cm wide rectangular sheet iron to cut a circle with the largest area. You can cut () at most. The area of each circle is（
Use a 15 cm long, 4 cm wide rectangular sheet iron to cut a circle with the largest area. You can cut () at most. The area of each circle is () square meters

Use a rectangular sheet of iron 15 cm long and 4 cm wide to cut a circle with the largest area. The maximum number of circles can be (3). The area of each circle is (12.56) square meters

Simple calculation, (11 2 / 13 + 13 2 / 11) / (5 / 13 + 5 / 11) = 23 / 25x16 / 17 + 16 / 25x2 / 17=
1.2+2.3+3.4+4.5+...+7.8+8.9 =

Molecule: 11 2 / 13 + 13 2 / 11 = 145 / 13 + 145 / 11 = 1595 / 143 + 1885 / 143 = 3480 / 143
Denominator: 5 / 13 + 5 / 11 = 55 / 143 + 65 / 143 = 120 / 143
The whole fraction: numerator / denominator = 3480 / 120 = 29

If a set of data x1, X2, X3 The variance is S2, then a set of new data 2x1 + 1, 2x2 + 1, 2x3 + 1 The variance of 2xn + 1 is______ ．

∵ a set of data x1, X2, X3 The variance is S2, a set of new data 2x1 + 1, 2x2 + 1, 2x3 + 1 The variance of 2xn + 1 is 22 × S2 = 4s2