Design algorithm to realize simple operation of one variable polynomial Basic requirements: (1) Input and build polynomial; (2) Output polynomial; (3) Polynomial addition (4) Polynomial subtraction Test data: （1）(2x+5x8－3.1x11)+(7－5x8+11x9)=(－3.1x11+11x9+2x+7) （2）(6x－3－x+4.4x2－1.2x9)－(―6x―3－+5.4x2－x2+7.8x15) =(―7.8x15―1.2x9+12x―3―x) （3）(1+x+x2+x3+x4+x5)+(―x3―x4)=(1+x+x2+x5) （4）(x+x3)+(―x―x3)=0 （5）(x+x100)+(x100+x200)=(x+2x100+x200) （6）(x+x2+x3)+0=x+x2+x3

Design algorithm to realize simple operation of one variable polynomial Basic requirements: (1) Input and build polynomial; (2) Output polynomial; (3) Polynomial addition (4) Polynomial subtraction Test data: （1）(2x+5x8－3.1x11)+(7－5x8+11x9)=(－3.1x11+11x9+2x+7) （2）(6x－3－x+4.4x2－1.2x9)－(―6x―3－+5.4x2－x2+7.8x15) =(―7.8x15―1.2x9+12x―3―x) （3）(1+x+x2+x3+x4+x5)+(―x3―x4)=(1+x+x2+x5) （4）(x+x3)+(―x―x3)=0 （5）(x+x100)+(x100+x200)=(x+2x100+x200) （6）(x+x2+x3)+0=x+x2+x3

Two stacks are used to store the read data and symbols, and an array is built to store the symbol priority
2 memory number and symbol. When encountering symbol, judge the priority (compared with the top layer of stack, if it is high priority) and calculate the number on both sides
3 put the result on the stack (delete the used data and symbols), repeat 2
The key is to determine the priority and store the number and symbol separately

Data structure (C language) uses single linked list to store one variable polynomials and realize the addition of two polynomials

#Include & lt; stdio. H & gt; & nbsp; & nbsp; # include & lt; stdlib. H & gt; & nbsp; & nbsp; # include & lt; malloc. H & gt; & nbsp; & nbsp; & nbsp; typedef & nbsp; int & nbsp; ElemType; & nbsp; & nbsp; & nbsp; & nbsp; / * statement of single linked list * / & nbsp; & nbsp; & nbsp; ty

What are the coefficients of polynomials

The whole polynomial has no coefficients
Only monomials have coefficients

Inequality solution of absolute value in high school
|3x+1|＜5
|2x+1|＞0
|3x+5|≤0
|6x+4|＞-1
Can use the interval method to write out better. 30 minutes. Urgent

|3x+1|＜5
-5

Is there any way that absolute value inequality can't be solved

In my opinion, the solution of absolute value inequality should be divided into different types. Different problems have different tricks. What I said is definitely not complete. I hope it can help you a lot. The most important thing is to remove the absolute value sign
(1) When there is only one sign of absolute value, such as | X-1|

The minimum value of the function y = | x-4 | + | X-6 |, is______ ．

The combination of number and shape: y = | x-4 | + | X-6 | = 10 − 2x & nbsp; & nbsp; & nbsp; & nbsp; (x ≤ 4) 2 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (4 < x < 6) 2x + 10 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (x ≥ 6) draw its image, as shown in the figure

(1) For any real number a, B, prove a ^ 2 + 3B ^ 2 ≥ 2B (a + b)
(2) For any real number AB, prove a ^ 2 + B ^ 2-2a-2 ≥ 0
(3) Given ABC positive integers, proving (with mean value theorem)
a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)≥6abc
（a^2+1）(b^2+1)(c^2+1)≥8abc
The second question is wrong. For any real number AB, prove a ^ 2 + B ^ 2-2A-2b + 2 ≥ 0

Two problems of solving absolute value inequality
Question 1: the absolute value of 2x-1 / 3 = 1 / 2

First question:
-1/3

Solving two absolute value inequalities
|X | is less than a (a is greater than 0)
|X | is greater than a (a is greater than 0)
It's better to write the solution

There is only one step to solve the problem. I can't go into more details unless we discuss it in categories. It's a big deal if we discuss it in categories
|x| < a
-a < x < a
|x| > a
X > A or x < - A

Solving inequalities with absolute values
The solution set of inequality [x + 3] - [X-2] > = 3 is? Xiaosheng doesn't understand this kind of inequality with absolute value and greater than or less than the number after it. Can the knower find a way of thinking in the process of solving it? I wish you happiness,

First set the range of X, and then go to the absolute value directly. As for solving piecewise function, this problem is as follows: when x ≤ - 3 - (x + 3) - (2-x) = - 5 ≥ 3, does not hold; when x ≤ - 3, the inequality has no solution; when - 3 ＜ x ≤ 2, (x + 3) - (2-x) = 2x + 1 ≥ 3, the solution is x ≥ 1; when - 3 ＜ x ≤ 2, we can know 1 ≤ x ≤ 2; when x ＞ 2, we can get x ≥ 1