1. If the solution of the equations {x + y = 3k, X-Y = 7K} about X, y satisfies the equation 2x + 3Y = 6, find the value of K 2, for X, y 1. If the solution of the equations {x + y = 3k, X-Y = 7K} about X, y satisfies the equation 2x + 3Y = 6, find the value of K 2. For X and y, we define a new operation "*": X * y = ax + by, where a and B are constants. On the right side of the equation are the usual addition and subtraction and multiplication operations. We know that 3 * 5 = 15, 4 * 7 = 28, and find 1 * 1 3. If the solution of the equations {3x + 2Y = P + 1,4x + 3Y = P-1} x ＞ y, the value range of P is obtained 4. It is known that the solution of the system of linear equations {2x + y = 6m, 3x-2y = 2m} with respect to X and Y satisfies the condition that one-third of x-one-fifth of y = 4, and the value of M is obtained

1. If the solution of the equations {x + y = 3k, X-Y = 7K} about X, y satisfies the equation 2x + 3Y = 6, find the value of K 2, for X, y 1. If the solution of the equations {x + y = 3k, X-Y = 7K} about X, y satisfies the equation 2x + 3Y = 6, find the value of K 2. For X and y, we define a new operation "*": X * y = ax + by, where a and B are constants. On the right side of the equation are the usual addition and subtraction and multiplication operations. We know that 3 * 5 = 15, 4 * 7 = 28, and find 1 * 1 3. If the solution of the equations {3x + 2Y = P + 1,4x + 3Y = P-1} x ＞ y, the value range of P is obtained 4. It is known that the solution of the system of linear equations {2x + y = 6m, 3x-2y = 2m} with respect to X and Y satisfies the condition that one-third of x-one-fifth of y = 4, and the value of M is obtained

1. If x + y = 3K is listed as (1), X-Y = 7K is listed as (2); if (1 + 2), 2x = 10K is obtained, that is, x = 5K, and y = - 2K is obtained by substituting (1); then k = 3 / 2 can be obtained by substituting X and Y into 2x + 3Y = 6

If a and B are the two real roots of the equation x & # 178; - 30x + 100 = 0, then LGA + LGB = how many) with the help of the gods,

2

Why (LGB / a) & 178; = (LGB + LGA) & 178;

No
The formula is LG (B / a) = LGB LGA
So it should be (LGB / a) & # 178; = (LGB LGA) & # 178;

If LGA and LGB are two real roots of the equation 2x2-4x + 1 = 0, then the value of AB is equal to ()
A. 2B. 12C. 100D. 10

∵ LGA, LGB are the two real roots of the equation 2x2-4x + 1 = 0. According to Weida's theorem, LGA + LGB = - − 42 = 2, ∵ AB = 100

Given the set a = {XLX & # 178; + (2 + P) x + 1 = 0, X ∈ r}, B = {x | x > 0} and a ∩ B = empty set, find the value range of real number p? I want to know - 4

1.A=∅
(2+p)²-4＜0
p²+4p

Given the set a = {XLX & # 178; - 2x-3 = 0}, B = {xlax-2 = 0}, find a ∪ B = a, find the value of real number a

A:(x-3)(x+1)=0;
X = 3 or x = - 1;
B：ax=2;
∵AUB=A;
(1) B is an empty set;
(2) B is not an empty set;
x=2/a;
2 / a = 3, then a = 2 / 3;
2 / a = - 1, then a = - 2;
The value of a is 0,2 / 3 or - 2;

It is known that a = {x | X & # 178; - 2x-3 ≤ 0, X ∈ r}, B = {x | X & # 178; - 2mx + M & # 178; - 4 ≤ 0, X ∈ R, m ∈ r}. If a ∩ B = [0,3], then
If a is included in CRB, find the value range of real number M

A = [- 1,3], B = [m-2, M + 2] can be easily obtained from the problem
Easy to get m = 2
2.. CRB = (- ∞, m-2) U (M + 2, + ∞), if a is contained in CRB
Then 3M + 2
The solution is m = (- ∞, - 3) U (5, + ∞)

The known set a = {y | y = x & # 178; - 3 / 2x + 1, X &; [3 / 4,2]}, B = {x | x + M & # 178; > = 1}“
 &; a "is a sufficient condition for" X &; B ", and the value range of real number m is obtained

Set a is the range of functions;
The symmetry axis of the function y = x ^ 2 - (3 / 2) x + 1 is x = 3 / 4, and the function is an increasing function in the domain of definition,
7/16≤y≤2
a=[7/16,2]
b=[1-m^2,+∞)
According to the meaning of the title:
A is the proper subset of B;
1-m^2≤7/16
m^2≥9/16
M ≥ 3 / 4, or m ≤ - 3 / 4

N={x||x-1/i|

It is known from the definition of circle (the set of points in the plane whose distance to a fixed point is equal to a fixed length)
|x-1/i|

The absolute value of (x + I) is less than root 2, and I is an imaginary unit

It is easy to know that the real part of complex x + I is x, (x ∈ R), and the imaginary part is 1
The module of the complex number is
|x+i|=√(x²+1)
From the question set | x + I | 2, it can be concluded that:
√(x²+1)＜√2
∴x²+1＜2
∴x²＜1
∴-1＜x＜1