Factorization of polynomial x ^ 2-4xy + 2Y ^ 2

Factorization of polynomial x ^ 2-4xy + 2Y ^ 2

Original formula = x ^ 2-4xy + 4Y ^ 2-2y ^ 2
=(x-2y)^2-(√2y)^2
=(x-2y+√2y)(x-2y-√2y)
I don't understand it
How to define unknowns in MATLAB, such as X,
The syntax is Syms X
Is to set a variable, otherwise matlab does not understand what x is
The polynomial x3-4x2y + 4xy2 is factorized and the result is______ .
x3-4x2y+4xy2，=x（x2-4xy+4y2），=x（x-2y）2．
What do Syms and int functions in MATLAB mean?
for instance:
syms x;
int(x)
int(x,0,1)
The results are as follows
ans =
x^2/2
ans =
1/2
Syms means to define a symbolic variable, which is different from those numerical variables and is used for formula simplification and calculation
Int means integral. The first one is indefinite integral. Of course, you can also calculate definite integral and write the upper and lower limits as in the second sentence. Of course, you can also integrate one of the variables. For example, you can use this command:
syms x z;
int(x/(1 + z^2),z)
result:
ans =
x*atan(z)
What else is the problem?
By factoring the polynomial X & # 178; - 4xy + 2Y & # 178;, we obtain——
x²-4xy+2y²=（x-2y-y√2）（x-2y+y√2）
I know the key to this answer is to put forward the answer of (x-2y-y √ 2) (x-2y + y √ 2) that y
Let z = x / y
Original formula = y ^ 2 (Z ^ 2-4z + 2)
=y^2(z^2-4z+2)
=y^2(z-2+√2)(z-2-√2)
=y^2(x/y-2+√2)(x/y-2-√2)
=(x-2y+y√2)(x-2y-y√2).
Missing row Vandermonde determinant
How to calculate the Vandermonde determinant with missing lines
Using the method of adding edges, you can add any line of the less Vandermonde determinant, and then add another column next to it. I'll write you the detailed process tomorrow. I have something to do today, so it's too late!
Today I will write you a detailed process:
For example, the determinant is as follows: (similar to Vandermonde determinant with missing row)
1 1 1 1
a b c d
a^2 b^2 c^2 d^2
a^4 b^4 c^4 d^4
We use the method of Canada Bank to solve this problem
After adding rows, the determinant becomes 5 rows and 5 columns, as follows:
1 1 1 1 1
a b c d x
a^2 b^2 c^2 d^2 x^2
a^3 b^3 c^3 d^3 x^3
a^4 b^4 c^4 d^4 x^4
This is the standard Vandermonde determinant
By using the determinant expansion rule and expanding according to the fifth column, the expansion formula is as follows:
A15 + (- a25) * x + A35 * x ^ 2 + (- D) * x ^ 3 + A55 * x ^ 4 [where a is an algebraic covalent and D is the value of the preceding fourth-order determinant]
According to the Vandermonde determinant formula, the value of the fifth order determinant is as follows:
(b-a)(c-a)(c-b)(d-a)(d-b)(d-c)(x-a)(x-b)(x-c)(x-d)
It is equal to the above expansion. What we need is the value of determinant D, so what we need to calculate is the coefficient of x ^ 3 in the expansion, so we get d =
(a+b+c+d)(b-a)(c-a)(c-b)(d-a)(d-b)(d-c)
If the absolute value of x-3 minus x + 4 is less than or equal to a, it holds for any x, and the value range of a is obtained
This problem is to find the maximum value of X - 3 minus x + 4
If 3 ＜ x, then │ x-3 │ - x + 4 = x-3-x-4 = - 7
If - 4 ＜ x ≤ 3, then │ x-3 │ - │ x + 4 = 3-x-x-4 = - 2x-1, because x ＞ - 4, so - 2x-1 ＜ 7
If x ≤ - 4, then │ x-3 │ - x + 4 = 3-x + X + 4 = 7
Therefore, the original maximum value is 7, which can be more vividly answered by connecting with the number axis, that is, x-3 is the distance from X to 3 on the number axis, x + 4 is the distance from X to - 4 on the number axis, and the maximum difference between the two distances is 7, that is, when x ≤ - 4
A is 7
How to find the sign inverse matrix of matrix by MATLAB
Let me give you an example
>>Syms a B C D (define variable)
>>A = [a, B; C, D] (definition matrix)
A =
[ a,b]
[ c,d]
>>Inv (a)
Ans = (result)
[ d/(a*d - b*c),-b/(a*d - b*c)]
[ -c/(a*d - b*c),a/(a*d - b*c)]
If P = | 1-2x | + | 1-3x | +, P = | 1-2x | +, P = | 1-3x | +, P = | 1-2x | +, P = |, P = |, P = |, P +|If the value of 1-9x | + | 1-10x | is constant, then this value is?
Please explain the calculation steps
When x is between positive and negative 1 / 10, ten absolute values take off the sign of absolute value, they all take 1 as positive and the term with X as negative, so the final result still has x term and can not be constant, so it is not in line with the meaning of the problem; when x is between positive and negative 1 / 10 and 1 / 9, after taking off the sign of absolute value, only the sign before 10x is positive and the term before 1 is negative
Matrix determinant
Let a be a matrix of order 3 | a | = 0.5, find the inverse of | (2a) - 5 (Adjoint of a)
|(2A)^(-1) － 5A* |
= |A||(2A)^(-1) － 5A*| /|A|
= 2 |A(2A)^(-1) － 5AA* |
= 2 |1/2(2A)(2A)^(-1) － 5|A|E |
= 2 |(1/2)E － (5/2)E |
= 2 |(-2)E |
= 2 (-2)^3
= -16