(3K square + 7K) + (4K square - 3K + 1) =? Urgent!

(3K square + 7K) + (4K square - 3K + 1) =? Urgent!

Merge congeners
3k²+7k+4k²-3k+1
=(3k²+4k²)+(7k-3k)+1
=7k²+4k+1
If you want to merge similar items, remove the brackets, add and subtract the same items, and you can easily get the answer 7K square + 4K + 1
What is the calculation method of determinant?
It is very important to make full use of the characteristics of determinant to simplify determinant. According to the characteristics of determinant, the second order reduction method transforms a row (column) into only one non-zero element by using the properties of determinant, and then expands it according to the row (column)
(3K & # 178; + 7K) + 1 / 2 + (4K & # 178; - 3K + 1)
(3K & # 178; + 7K) + 1 / 2 + (4K & # 178; - 3K + 1)
=(3k²+4k²)+(7k-3k)+(1/2+1)
=7k²+4k+3/2
Calculation method of determinant~
Is it the property of triangle determinant?
Or all the ways to find determinants?
If the latter, leave email, send you a reference
Please ask
It is known that the order of the n-th power of the monomial-2a is the same as that of the 3-th power of the polynomial x + the 3-th power of 4x multiplied by the 4-th power of Y - the 5-th power of 2Y
What is n? How many times is the number of this single item?
The highest degree of a polynomial is 3 + 4 = 7
So n = 7
The number of monomials is seven
How to calculate this determinant?
x a b c d
a x b c d
a b x c d
a b c x d
a b c d x
Add columns 2, 3, 4, 5 to column 1
The first column presents x + A + B + C + D
Add column 1 by - A to column 2
Add column 1 by - B to column 3
Add column 1 by - C to column 4
Add column 1 by - D to column 5
The determinant is reduced to the lower triangular form
D = (x+a+b+c+d)(x-a)(x-b)(x-c)(x-d)
If the difference between a polynomial a minus the square of 4x - 3Y + 5 is 2Y + 6, then a =?
A-((4x)^2-3y+5)=2y+6
A=16x^2+5y+1
Determinant calculation
4 3 0 0
1 4 3 0
0 1 4 3
0 0 1 4
One of the answers is 4300 = 0 - 13 - 130 = 0 04039 = 0 000 - 121
1 4 3 0 1 4 3 0 1 4 3 0 1 4 3 0
0 1 4 3 0 1 4 3 0 1 4 3 0 1 4 3
0 0 1 4 0 0 1 4 0 0 1 4 0 0 1 4
How can I wait for the next two steps? It was a very simple question, but I didn't understand the last two steps of this algorithm,
The last two steps are wrong
4 3 0 0 =0 -13 -13 0 = 0 0 39 39 = 0 0 0 -117
1 4 3 0 1 4 3 0 1 4 3 0 1 4 3 0
0 1 4 3 0 1 4 3 0 1 4 3 0 1 4 3
0 0 1 4 0 0 1 4 0 0 1 4 0 0 1 4
The final result should be 117
If the value of polynomial (2x & # 178; - 3x-ay + 6) - (2bx & # 178; - 3x + 3y-1) has nothing to do with the value of X. y, find the value of (a + b) (a-b)
(2x²-3x-ay+6)-(2bx²-3x+3y-1)
=2x²-3x-ay+6-2bx²+3x-3y+1
=2(1-b)x²-（a+3）y+7
Since the value of polynomial (2x & # 178; - 3x-ay + 6) - (2bx & # 178; - 3x + 3y-1) has nothing to do with the value of x.y,
So 1-B = 0, a + 3 = 0
b=1,a=-3
(a+b)(a-b)=a²-b²=（-3）²-1²=9-1=8
The original formula = (2-2b) x & # 178; - (a + 3) y, the unknown coefficient should be 0, then B = 1, a = - 3, the final result is 8!
How about this determinant
1 2 3 ...n-1 n
2 3 4 ...n 1
3 4 ...1 2
.
.
.
n 1 2 ...n-2 n-1
How much should this determinant cost
Help···
Have you learned determinants? If you have, you should know the equivalent change of determinants. Use properties to transform back and forth. I went to sleep. From back to front, subtract the previous line 1 23... N-1 N1 1 1... 1-n1 1 1 1... 1-N 1... 1 1-N 1 1 1-N 1... 1 1-N 1-N 1 1... 1 or