(K-3) x ^ (K-2) + x power + KX + 1 / 2 = O is a quadratic equation of one variable about X, then the value of K is

(K-3) x ^ (K-2) + x power + KX + 1 / 2 = O is a quadratic equation of one variable about X, then the value of K is

Because it is a quadratic equation of one variable
So K-2 is a nonnegative integer less than or equal to 2
When K-2 = 0, the original formula = x ^ 2 + 2x-0.5 = 0
When K-2 = 1, the original formula = x ^ 2 + 3x + 0.5 = 0
When K-2 = 2, the original formula = x ^ 2 + 2x + 0.25 = 0
So K is 2, 3, or 4

The solution of ternary linear equation: x + 2Z = 3,2x + y = 2,2y + Z = 7
We should use the substitution method,

1.x+2z=3 ①
2x+y=2 ②
2y+z=7 ③
By using 1 × 2 - 2, we get 4z-y = 4, 4
In this paper, we use 4 × 2 + 3, 9z = 15, z = 5 / 3
Substituting z = 5 / 3 into formula 3, we get 2Y + 5 / 3 = 7, y = 8 / 3
Substituting y = 8 / 3 into 2, we get 2x + 8 / 3 = 2, x = - 1 / 3

If x / 3 = Y / 4 = Z / 5, find the value of X + y + Z / 3x-2y + Z
There is a big reward of 100

Let X / 3 = Y / 4 = Z / 5 = M
Then x = 3M, y = 4m, z = 5m
Then x + y + Z / 3x-2y + Z = (3m + 4m + 5m) / (9m-8m + 5m) = 12 / 6 = 2

Given X / y = 2 / 3, find the value of 3x-y / x + 2Y

x=2/3y
2y-y/2/3y+2y=3/8

Given x ^ 2-2 √ 3x + y ^ 2 + 2Y + 4 = 0, find the value of Y / X
The answer is - 3 / 3
Make the answer clear

x²-2√3x+3+y²+2y+1=0
(x-√3)²+(y+1)²=0
If the sum of the squares is 0, it is equal to 0
So x - √ 3 = 0, y + 1 = 0
x=√3,y=-1
So y / x = - √ 3 / 3

Given | x | = 2, | y | = 3, find the value of 3x + 2Y

∵|X|=2,|Y|=3
∴X=±2,Y=±3
Ψ 3x + 2Y = 6 + 6 = 12 or - 6 + 6 = 0 or 6-6 = 0 or - 6-6 = - 12

Let β 1. β 2 be the two solution vectors of the non-homogeneous linear equation system AX = B. let any of the following vectors be the solution of the equation system, then it is ()
A.β1+β2.B.1/5(3β+2β) C.1/2(β1+2β2） D.β1-β2

Choose B
A[1/5(3β1+2β2)]=3/5Aβ1+2/5Aβ2=3/5b+2/5b=b
So 1 / 5 (3 β 1 + 2 β 2) is the solution of the system

Let A1 and A2 be the two solution vectors of AX = B, then a ((2A1 + 3a2) / 5) =?

It is known that Aa1 = B, aa2 = B
So a ((2A1 + 3a2) / 5) = (2aa1 + 3aa2) / 5 = (2B + 3b) / 5 = B
That is, (2A1 + 3a2) / 5 is still the solution of AX = B

Let α 1, α 2 The vector β is not the solution of AX = 0, that is, a β ≠ 0 β + α t is linearly independent

Suppose there is a set of constants K, K1 In this way: (K + Ti = 1Ki) β = Ti = 1 (− Ki) α I α t is a system of homogeneous linear equations ax

Let a be a matrix of order n. for the homogeneous linear equation system AX = 0, if the sum of elements in each row of a is 0 and R (a) = n-1, then the general solution of the equation system is?, if every n-dimensional column vector is the solution of the equation system, then R (a) =?

It is obvious that (1,1,..., 1) ^ t is the nonzero solution of AX = 0. Substitute R (a) = n-1 into the formula
Number of solution vectors = number of unknowns rank of coefficient matrix = n - (n-1) = 1
So the equation has only one solution vector, so the general solution is x = K (1,1,..., 1) ^ t, where k is an arbitrary constant
If every n-dimensional column vector is the solution of the system of equations, it means that the solution vector can describe every vector in the whole space, and we know that only a group of vectors whose number and space dimension are equal and linearly independent can do this, such as the XYZ coordinate in the 3-dimensional space, so the equation has n solution vectors, and it is easy to get the rank of the matrix as 0 by substituting it into my above formula again