If the solution of equation 2-x = 2 (x-1) is the same as that of equation A-3 (x-1 / 3) = 2, then a=

If the solution of equation 2-x = 2 (x-1) is the same as that of equation A-3 (x-1 / 3) = 2, then a=

2-X=2(x-1)
2-x=2x-2
3x=4
x=4/3
Substituting A-3 (x-1 / 3) = 2
a-3(4/3-1/3)=2
a-3=2
a=5

Given X & # 178; - 3 = 0, find the value of the algebraic formula x (x + 1) &# 178; - x (X & # 178; + x) - X-9

Because X & # 178; - 3 = 0, X & # 178; = 3x (x + 1) &# 178; - x (X & # 178; + x) - X - 9 = x (x + 1) (x + 1) - X & # 178; (x + 1) - X - 9 = x (x + 1) - X - 9 = x & # 178; - 9 = 3 - 9 = - 6

If the polynomial x ^ 4 + (A-1) x ^ 3 + 5x ^ 2 + (B + 3) X-1 does not contain x ^ 3 and X terms, find the value of a + B

a-1=0;
b+3=0;
a=1;
b=-3;
a+b=1-3=-2;
If you don't understand this question, you can ask,

At the sports meeting, the life class committee took 20 yuan to the supermarket to buy x bottles of juice and Y bottles of yoghurt for the athletes. It is known that each bottle of juice is 2 yuan and each bottle of yoghurt is 3 yuan. When the money is just used up, there is a total purchase plan______ Species

From the meaning of the title, we can get 2x + 3Y = 20, X ≥ 0, y ≥ 0, and X and y are integers. When y = 0, x = 10, when y = 1, x = 8.5, when y = 2, x = 7, when y = 3, X = 5.5, when y = 4, x = 4, when y = 5, x = 2.5, when y = 6, x = 2, x = 2, x = 2, x = 4, y = 4, y = 4, y = 4, y = 4, y = 4, y = 5, y = 4, y = 4, y = 5, y = 4, y = 4, y = 5, y = 2, y = 6 There are 4 purchase options. So the answer is: 4

A is an invertible matrix of order n. It is proved that there exists a positive definite matrix s and an orthogonal matrix P such that a = PS

Let a 'be the transpose of a, and consider B = a'a. then B is a positive definite matrix
It can be proved that there exists a positive definite matrix s such that B = S & #
If P = as ^ (- 1), then p '= (s') ^ (- 1) a' = s ^ (- 1) a '
P'P = S^(-1)A'AS^(-1) = E.
Then p is an orthogonal matrix. A = PS satisfies the requirement

Factorization of P (X-Y) + Q (Y-X)

p(x-y)+q(y-x)
=p(x-y)-q(x-y)
=(p-q)(x-y)

The value range of the letter A in the fraction A / (3a & # 178; + 8a-3) is
If the value of fraction (- 2x & # 178; + 7x-5) / (X & # 178; + X-2) is zero, then the value of X is zero
The equation 2T + 1 / 10 = - 5 / 2T & # 178 is solved by the collocation method;

3a²+8a-3≠0
（3a-1）（a＋3）≠0
A ≠ 1 / 3 and a ≠ - 3
The value of fraction (- 2x & # 178; + 7x-5) / (X & # 178; + X-2) is zero
（-2x²+7x-5）=0,
（x-1）（2x-5）=0
X = 1 or x = 5 / 2
but
（x²+x-2）≠0
（x-1）（x＋2）≠0
x≠1
therefore
x=5/2

If f (x) = INX + (x-a) ^ 2, a is a constant
If x = 1, f (x) gets the extremum, find the value of a, and find the monotone increasing interval of F (x)

f'(x)=1/x+2(x-a)
f'(1)=1+2-2a=0
a=1.5
f'(x)=1/x+2x-3
=(2x^2-3x+1)/x
=(2x-1)(x-1)/x
0

In order to know the inverse scale function, the image with y equal to K / X passes through point a (- 2,3)
Is there any other intersection point for the image passing through the positive ratio function y = K / X of point a? If so, please explain the reason

（2,－3）（－2,3）

If the product of five rational numbers is negative, what is the number of negative factors among them?

Singular (1,3,5)