Matlab polynomial substitution Q=14*t*u*w*x^4+14*z*v*y*x^4+7*z*t^2*x^4+7*w^2*v*x^4+7*y^2*w*x^4+7*z^2*u*x^4+7*u^2*y*x^4+7*t*v^2*x^4 How do I calculate the expression of Q when t = 0, z = 0, v = 0

Matlab polynomial substitution Q=14*t*u*w*x^4+14*z*v*y*x^4+7*z*t^2*x^4+7*w^2*v*x^4+7*y^2*w*x^4+7*z^2*u*x^4+7*u^2*y*x^4+7*t*v^2*x^4 How do I calculate the expression of Q when t = 0, z = 0, v = 0

This is to give an expression, known t = 0, z = 0, v = 0, the rest of the U W x y as variables, so programming
t=0;z=0;v=0;
syms u w x y ;
Q=14*t*u*w*x^4+14*z*v*y*x^4+7*z*t^2*x^4+7*w^2*v*x^4+7*y^2*w*x^4+7*z^2*u*x^4+7*u^2*y*x^4+7*t*v^2*x^4;
factor(Q)
After the implementation of the program, the results are as follows:
ans =
7*x^4*y*(u^2 + w*y)

Find the graph of y = | x + 1| - 2| - 3|

This is a piecewise function, 1, when X

1.1/x|x/(x+1)|

1, x > 1 or X

How to solve the absolute value inequality with [parameter]?
|2x-1|＞ax+3
How to solve this system of equations? Why can't it be substituted into the formula | f (x) | g (x) f (x) > G (x) or F (x) < - G (x)? Doesn't it mean that we don't need to discuss whether the right side is greater than 0 to solve this system of equations? Why is it different with one more parameter?

This kind of equation is to first discuss whether the right term is greater than 0. If it is less than 0, find out the range of A. obviously, no matter how many values of X are in this range, the equation holds. If it is greater than oh, it is OK to substitute the formula | f (x) | g (x) f (x) > G (x) or F (x) ＜ - G (x) to solve the specific value

How to discuss the solution set of absolute value parameter inequality
Please explain

Generally speaking
For the inequality f (x) > | g (x)|
Then, when - G (x) 0, x 3, and then use the solution x to find the intersection of this set

Three questions on absolute value inequality
Three questions on absolute value inequality
1. We know that the solution set of a|x-b|is_________
3. A = {x | x-a | 3}, and a and B = R, find the value range of real number a

Two problems of absolute value inequality
|x-a|

The second: given H > 0, proposition a is that two real numbers a and B satisfy A-B < 2h, and proposition B is that two real numbers a and B satisfy A-1 < h and B-1 < H,
Then a is a necessary and insufficient condition for B
A cannot push B: let a = 2 + H, B = 2 + H, then a is satisfied and B is not satisfied
B can push a | A-B | = | (A-1) - (B-1) | ≤| A-1 | + | B-1 | < 2H

A problem of absolute value inequality
If 2-m and | m | - 3 are different signs, then the value range of M is?

(2-m) (| m | - 3)

A question about absolute value inequality
If the inequality | X-2 | - | x + 1 | > m about X has no solution, find the value range of M

|X-2 | - | x + 1 | > m has no solution
When XM, 3 > m (1)
When - 1 ≤ x ≤ 2, the inequality is: 2-x - (x + 1) > m, 1-2x > m (2) X2, the inequality is: x-2-x-1 > m, - 3 > m (3)
So (1) (2) (3) are empty
That is, the value range of M is: m ≥ 3

Exercises for solving some absolute value inequalities
For example, exercises like | x + 3 | - | x-4 | + | X-5 |

|x+a|-|x-b|+|x-c|
We can get x = - A, B, C
Compare the size of - A, B, C
If - a > b > C
The discussion is divided into situations
1.x