Find the range of values in which the solutions of the equations x-2y = 7m-8, 3x-2y = 17m-8 about X and y are all positive numbers

Find the range of values in which the solutions of the equations x-2y = 7m-8, 3x-2y = 17m-8 about X and y are all positive numbers

Let m be a constant to solve the equations
x=5m y=-m+4
If the meaning of the title is x > 0, Y > 0, that is - M + 4 > 0, then M0
So 0

How to find the degree and coefficient of polynomial?

The coefficient, for example, 3x + 5Y, is 3 and 5, and the degree, for example, 3x, is 2

The method of determining the coefficient and degree of polynomial term

A:
The coefficient and degree of polynomial, the key is to look at the monomial
Let's look at the number of times each monomial is used
The degree of the binomial with the highest degree is the degree of the polynomial
There is no coefficient in polynomial, only degree and several terms
For example, five times trinomial

Let f (x) have f (x1 + x2) = f (x1) + F (x2) for any real number x1, X2, and f '(0) = 1. Prove that f' (x) = f (x)

When f '(x) = f (x), only f (x) = e ^ X
It is obvious that if f (x1 + x2) = f (x1) + F (x2), the function of a degree does not satisfy the condition of F '(x) = f (x)
If f (x1 + x2) = f (x1) * f (x2) and f '(0) = 1, then f (x) = e ^ X

It is proved that the function f (x) = - 2x + 1 is a decreasing function on R. it is proved that let X1 and X2 be any two real numbers on R. and why is X10 not f (x1) - f (x2)

Because f (x1) - f (x2) = (- 2x1 + 1) - (- 2x2 + 1) = 2 * (x2-x1) is in x1x1, that is, x2-x1 > 0, then multiply 2 left and right to get 2 * (x2-x1) > 0
That is, f (x1) - f (x2) = 2 * (x2-x1) > 0. Then f (x1) - f (x2) > 0

It is known that f (x) is an increasing function on real number r, and f (x) = f (x) - f (2-x) if f (x1) + F (x2) > 0, it is proved that X1 + x2 > 2

The following is to the contrary
If X1 + x2

It is known that: F (x) = loga ^ x (a > 0), if X1 and X2 are positive real numbers, compare the sizes of 1 / 2 {f (x1) + F (x2)} and f [(x1 + x2) / 2], and prove that]

F (x1) + F (x2) = loga (x1 * x2) so 1 / 2 [f (x1) + F (x2)] = 1 / 2 * loga (x1 * x2) = loga [√ (x1 * x2)] f [(x1 + x2) / 2] = loga [(x1 + x2) / 2] from the mean inequality (x1 + x2) / 2 > = √ (x1 * x2), so if 01, f (x) is an increasing function, loga [√ (x1 * x2)]

Whether there is a real number k, so that (2x1-x2) (x1-2x2) = 3 / 2 holds, whether it exists, explain the reason
It is known that X1 and X2 are two real roots of the quadratic equation 4kx2-4kx + K + 1 = 0

Step1：∵4kx2-4kx+k+1=0,
According to Weida's theorem,
x1x2=（k+1）/4k,x1+x2=-(-4k/4k)=1
(2x1-x2) (x1-2x2) = 2 (x1 + x2) ^ 2-9x1x2 ("^ 2" means "square")
=2-9(k+1)/4k
Step 2: let 2-9 (K + 1) / 4K = 3 / 2,
The solution is k = - 9 / 2
P. S. would you mind offering some reward points?

Let p be a real number, there are two different real roots x1, X2 about x ^ 2-3px-p = 0. (1) prove: 3px1 + x2 ^ 2-P > 0
(2) Find the minimum value of u = (P ^ 2 / (3px1 + x2 ^ 2 + 3P)) + ((3px2 + X1 ^ 2 + 3P) / P ^ 2)

It is easy to know that X1 + x2 = 3P, X1 * x2 = - p3px1 + x2 ^ 2-P = (x1 + x2) X1 + x2 ^ 2 + X1 * x2 = (x1 + x2) ^ 2 = 9p ^ 2 because X1 is not equal to X2, so p is not equal to 0, so 3px1 + x2 ^ 2-P > 03px1 + x2 ^ 2 + 3P = (x1 + x2) X1 + x2 ^ 2-3x1x2 = X1 ^ 2 + x2 ^ 2-2x1x2 = (x1-x2) ^ 23px2 + X1 ^ 2 + 3P = (x1 + x2) x2 + X1 ^ 2

The absolute value of X1 = X2, why is X1 equal to plus or minus x2

Because the absolute value is positive, but if X1 is negative, just add the absolute value to make it positive
So there are + X1 and - x2