Question 21: How to do it? How to do Question 21?

There are n-1 zeros, respectively in the interval [1,2],[2,3].n-1, n].

f'The point at which (x)=0 is the extreme point of the function f(x), f(1)=f(2)=0, and f(x) is continuous, so there must be an extreme point on [1,2] and only one extreme point. f(x)=0 has n solutions, f'(x)=0 has n-1 extreme points.

The interval of other extreme points and so on.

f(x) is the nth order equation of x, and the equation can also be written.

After the derivative f'(x) is the (n-1) order equation of x, then f'(x)=0 has n-1 solutions.

Let p(x1,y1).

When a=0, x1=x0, y1=(-2c b)-y0; When b=0, x1=(-2c a)-x0 and y1=y0 can be solved

When ab is not equal to 0.

Because it is symmetrical, the midpoint of the line segment is on a straight line: a(x1+x0) 2+b(y1+y0) 2+c=0

Vertical: (y1-y0) (x1-x0)*(a b)=-1 solves x1=-(2ca+aby0-x0b 2y0) (a 2+b 2y0), y1=(x1-x0)b a+y0

Using the diagonal rule, the solution is as follows:

Isn't there a process below?

Use the elementary transformation method below.

Can you be specific?

Questions. <>

OK Wait a minute. At the beginning, as soon as Mei Lanfang stood at such a high place, she was very panicked, and her waist was sore and her legs hurt after standing for a while. In order to practice excellent kung fu, Mei Lanfang gritted her teeth and persevered, and even her legs were swollen.

In winter, he poured a small Xiaoice rink himself, stepped on stilts, and ran on the ice rink. The smooth ice, not to mention stepping on stilts, even if you walk on it, you will inevitably have to fall. Mei Lanfang's body was often bruised and purple.

Every time he fell, he immediately got up and continued to practice.

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Questions. Brief summary, understand?

Mei Lanfang stood on a high place and practiced Mei Lanfang practiced stilts on the ice.

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Answer: g(x)=2x-a+1 x;

To make f(x) a monotonically increasing function constant g(x) 0

It is known from the mean inequality.

g(x)=2x-a+1/x≥2×sqrt(2x*1/x)-a=2×sqrt(2)-a≥0;

Therefore, a takes a maximum of 2 sqrt(2).

2) f(x) is constant on x>1, and f(1) 0

i.e. 1-a 0

a≤1.From (1), we know that when the derivative function of a 1, f(x) is everstable at zero, so f(x) increases monotonically, and there is f(x) f(1) 0

Therefore: a 1

Derivative f'(x)=2x 2-2a=2(x 2-a) 0<=x 2<1 when x belongs to (-1,1).

The discussion is as follows: 1) a<0

Obviously, f'(x) >0 is constant, the function is monotonically increasing, and there is no extremum.

2) When 0<=a<1, x 2-a=0 has 2 real solutions, x = a

The function is incremented at (-1,-a) and (a,1); There are two extrema decreasing at (- a, a), the maximum f(- a).

The minimum value f(a).

3) A>1.

Obviously, f'(x) <0 is constant, and the function is monotonically decreasing, and there is no extremum.

an=sn-1, except a1, a piecewise function can be obtained, and n=8 is easily obtained

21.(1) 1325 (1-3 5) = 1325 2 5 = 530 (km).

530 88 and 1 3 = 6 (hours).

A: There are 530 kilometers left, and it will take 6 hours to reach Tianjin.

2) 88 and 1 3 6 (1-3 5) = 530 2 5 = 1325 (km).

A: The railway from Shanghai to Tianjin is 1,325 kilometers long.