Fourth order determinant first row 1 + x 1 1 second row 1 1-x 1 third row 1 1 + x 1 fourth row 1-y I don't know the fourth-order determinant in detail

Fourth order determinant first row 1 + x 1 1 second row 1 1-x 1 third row 1 1 + x 1 fourth row 1-y I don't know the fourth-order determinant in detail

1+x 1 1 1
1 1-x 1 1
1 1 1+x 1
1 1 1 1-y
Start with the fourth line and subtract the first line from each line
1+x 1 1 1
-x -x 0 0
0 x x 0
0 0 -x -y
First and second lines
1 1-x 1 1
-x -x 0 0
0 x x 0
0 0 -x -y
The second line plus x times the first line
1 1-x 1 1
0 -x^2 x x
0 x x 0
0 0 -x -y
Expand by the first column,
-x^2 x x
x x 0
0 -x -y
-x^2[-xy-0]-x[-xy+x^2]
The first row of a fourth-order determinant is x 1 1, the second row is 1 x 1, the third row is 1 x 1, and the fourth row is 1 x 1
All the roots of the equation need specific steps to solve the problem, because this is a linear algebra management class, and it doesn't need to be complicated,
Add columns 2, 3 and 4 to column 1
Line 2, 3, 4 minus line 1
Transformation of determinant into upper triangular determinant
D = (x+3)((x-1)^3
So x = 1 or - 3
The first row x a, the second row a x a, the third row a x a, the fourth row a x
First line x a
The second line a x a
The third line a x a
The fourth line a x
C1 + C2 + C3 + C4 means that columns 2,3,4 are added to column 1
then
Ri-r1, I = 2,3,4, i.e. line 2,3,4 minus line 1
Determinant is transformed into upper triangular determinant
x+3a a a a
0 x-a 0 0
0 0 x-a 0
0 0 0 x-a
= (x+3a)(x-a)^3.
A line x a
zilou
1 / 2 (4x + 2Y) - 2 (X-Y) + 3x simplification!
1 / 2 (4x + 2Y) - 2 (X-Y) + 3x
=2X+Y-2X+2Y+3X
=3X+3Y
The graph of a function y = KX + B (K ≠ 0) passes through points (3,3) and (1, - 1). Find its functional relation
According to the meaning of the problem: 3K + B = 3K + B = − 1, the solution: k = 2B = − 3, then the analytic formula of the function is: y = 2x-3
The solution of inequality: log2 (x2-x-2) > log2 (2x-2)
The original inequality is transformed into x2 − x − 2 ﹥ 0x − 1 ﹥ 0x2 − x − 2 ﹥ 2x − 2 (4 points) ﹥ x − 2 (x + 1) ﹥ 0x − 1 ﹥ 0x2 − 3x ﹥ 0 (8 points) ﹥ x ﹥ 2x ﹤ 0 or X ﹥ 3 ﹥ the solution set of the original inequality is: {x | x ﹥ 3} (12 points)
It is known that the solution set of inequality (4a + 3b) x greater than B-A is x less than 4 / 9, and the solution set of ax greater than B is obtained
It is known that the solution set of inequality (4a + 3b) x greater than B-A is x less than 4 / 9, and the solution set of ax greater than B is obtained
Because the solution set of (4a + 3b) x > B-A is XB / a = - 25 / 3, so the solution set is x > - 25 / 3
Slope formula
Let the inclination angle of the straight line be α and the slope be k = Tan α = Y / X
Let the known point be (a, b) and the unknown point be (x, y)
k=(y-b)/(x-a)
Derivative: the derivative value of a point on the curve is the slope of the tangent of the point on the curve
Abscissa divided by ordinate~~
The equation AX + by + C = 0, when B ≠ 0, the slope is - A / b
3x+2y/4=2x+y+2/5=4x+5y/3
(3x+2y)/4=(2x+y+2)/5=(4x+5y)/3
It can be divided into (3x + 2Y) / 4 = (2x + y + 2) / 5 and (2x + y + 2) / 5 = (4x + 5Y) / 3
Diagonal multiplication
(3x+2y)/4=(2x+y+2)/5
5(3x+2y)=4(2x+y+2)
15x+10y=8x+4y+8
7x+6y=8 (1)
(2x+y+2)/5=(4x+5y)/3
3(2x+y+2)=5(4x+5y)
6x+3y+6=20x+25y
14x+22y=6
7x+11y=3 (2)
(2)-(1)
5y=-5
y=-1
Substituting (2), 7x-11 = 3
X=2
If you have a problem with your expression, / 4, / 5 / 3, where do you put them? Just find out how much x is equal to how much y of an unknown number from two equations, and then bring them into another formula. This is very simple
If the image of a linear function y = KX + B passes through P (3, - 2) and Q (- 1,2), then its expression
By substituting P and Q into the analytical expression, we get the following results
3k+b=-2
-k+b=2
Solution
k=-1,b=1
So y = - x + 1
Substituting two points into y = KX + B to get - 2 = 3K + B, 2 = - K + B, the solution is k = - 1, B = 1. So the expression is y = - x + 1
Substituting x = 3, y = - 2, x = - 1, y = 2 into y = KX + B
-2=3k+b
2=-k+b
k=-1
B=1
Y = - x + 1