Bivariate quadratic equation Suppose a system of equations of two variables, {X1 = 1, X2 = 2 y1=6 y2=3

Bivariate quadratic equation Suppose a system of equations of two variables, {X1 = 1, X2 = 2 y1=6 y2=3

(x-1)(y-3)=0
(x-2)(y-6)=0
xy=6
y=-3x+9
Or it can be constructed as in the message
But he wrote the opposite. Should be:
(x-1)(y-3)=0
(x-2)(y-6)=0
Given that the definition domain of function y = LG (x ^ 2-2x-3) is a, the solution set of inequality | X-1 | ≥ a (a > 0) is B, and a is included in B, what is the value range of a?
I hope it can be faster and make more money
The definition domain of function is the solution set of inequality X & # 178; - 2x-3 = (x + 1) * (x-3) > 0, so a = (- ∞, - 1) ∪ (3, + ∞). The solution set of inequality X-1 | ≥ a (a > 0) is: B = (- ∞, 1-A) ∪ (1 + A, + ∞). It is also known that a is contained in B, so 1 + a < 3 and a > 0. Then 0 < a < 2
Let u = R, a = {x ∈ R | a ≤ x ≤ 2}, B = {x ∈ R | 2x + 1 ≤ x + 3, and 3x ≥ 2}. (1) if a = 1, find a ∪ B, (∁ UA) ∩ B; (2) if B ⊆ a, find the value range of real number a
(1) If a = 1, if a = 1, then a = {{x | 1 ≤ x ≤ 2}, B = {x | x ≤ 2, and X ≥ 23} = {x | 23 ≤ x ≤ 2}, if a = 1, then a leta ∪ B = {x | 1 ≤ x ≤ 2}, B = {{x | 1 ≤ 1 ≤ x ≤ 2}, B = {x {x | 1 ≤ 1 ≤ x ≤ x ≤ 2}, B = {{{x ∁ UA) \\\\\\\\\\8705 \\\\\\inthis paper, we present a new method to solve the problem of} = {x | 23 ≤ x ＜ 1}
What's the difference between a bivariate equation and a bivariate equation?
For example, differences in form and concept, etc~
I forgot~
You're wrong. It's not a quadratic equation with one variable. It's a quadratic equation with two variables. Zhang ruofeng ddsncwj and Dian Dian are very big. They are very good. I just chose one whose answer is close to myself. Thank you very much for your reply!
·The quadratic equation with one variable has only one unknown number, and the power of the unknown number is one
General formula: ax + B = 0 (a is not equal to 0)
·Bivariate linear equation has two unknowns, and the power of the unknowns is one. Generally, it is composed of two bivariate linear equations to solve the unknowns x, y
Given that f (x) is a monotone increasing function defined in (0, + ∞), f (XY) = f (x) + F (y), and f (2) = 1, then the solution set of the inequality f (x) + F (x-3) ≤ 2 is obtained
I wrote x > 0, x-3 > 0, X (x-3) ≤ 4. Who can explain why f [x (x-3)] ≤ f (4) so as to get x (x-3) ≤ 4,
The idea is that it should be the solution set of F (x) + F (x-3) ≤ 2, so we need to merge "f (x) + F (x-3)";
Naturally, we use this condition: F (XY) = f (x) + F (y),
Then f [x (x-3)] = f (x) + F (x-3) ≤ 2
The problem becomes the set of solutions f [x (x-3)] ≤ 2,
Then we use f (2) = 1 and f (XY) = f (x) + F (y) to obtain
f（2*2）=f（2）+f（2）=2
Then the title becomes f [x (x-3)] ≤ f (4)
Because f (x) is a monotone increasing function defined at (0, + ∞)
There is x (x-3) ≤ 4
It should be clear
f(x)+f(x-3)≤2
f(x(x-3))≤2=2f(2)=f(4)
Because f (x) is a monotone increasing function defined at (0, + ∞), X (x-3) ≤ 4
because
F (x) is a monotone increasing function defined in (0, + ∞), so from F [x (x-3)] ≤ f (4), X (x-3) ≤ 4 is obtained
It's a function defined from 0 to + ∞, and it's monotonic increasing.
The complete set u = R, a = {x ∈ R | a + 1 + X ＞ 0}, the solution set of inequality {2x + 1 ≥ x-3, 3x + 2 ＜ o} is B
In order to make every x value in the set a satisfy at least one of the inequalities "1 < x < 3" and "x > 4 or x < 2", the value range of real number a is calculated
By satisfying at least one of the inequalities "1 ＜ x ＜ 3" and "x ＞ 4 or X ＜ 2" for each x value in the set B and a, we can get the range of X as [- 4,3) or (4, + ∞). Then x in the set a can be expressed as x > - A-1. Obviously, the range of X is monotonic, so the maximum range of X is (4, ∞), so the range of - A-1 is greater than or equal to 4, so a ∈ (- ∞, 5]
A problem of bivariate quadratic equation
It is known that the system of equations y ^ = NX, y = 2x = m (where m, n are not 0) has a real solution. Try to determine the value of M / n
Y ^ = (2x) ^ = 4xx n = 4x M = 2x n (that is 4x) / M (that is 2x) = 1 / 2
Let f (x) be a monotone increasing function on the definition (0, + ∞), and f (XY) = f (x) + F (y) and f (2) = 1 for any X and Y in the domain of definition. The range of values of X for which the inequality f (x) + F (x-3) ≤ 2 holds is obtained
f(xy)=f(x)+f(y)
f(4)=f(2×2)=f(2)+f(2)=1+1=2
f(x)+f(x-3)≤2
f(x (x-3))≤2＝f(4)
And f (x) is a monotone increasing function on the definition (0, + ∞)
X>0
And x-3 > 0
And 0
If the equation 3x & # 178; + MX = 2x-1 contains no linear term of X, then M=
Then (m-2) x = 0
So m = 2
The equation 3x & # 178; + MX = 2x-1 contains no linear term of X
M=2
By solving the binary linear equations 8x + 6y = 36x − 4Y = 5, y = ()
A. -112B. -217C. -234D. -1134
(1) In this way, we choose D