When k is the value, the solution of the equations 4x + 3Y = 60, KX + (K + 2) y = 60 satisfies x > y > 0

When k is the value, the solution of the equations 4x + 3Y = 60, KX + (K + 2) y = 60 satisfies x > y > 0

4X + 3Y = 60 (1) KX + (K + 2) y = 60 (2) we can get 4kx + 3ky = 60K (3) 4kx + (4K + 8) y = 240 (4) (3) (4) by subtracting (3k-4k-8) y = 60k-240, we can get y = 60 (4-K) / (K + 8) and substitute (1) to get x = 60 (k-1) / (K + 8) known x > y > 060 (k-1) / (K + 8) > 60 (4-K) / (K + 8) > 0
Because: x > y > 0
So: 4-ky > 0
So, we have to
k>2.5
From the question: 4x + 3Y = KX + (K + 2) y
(4-k)x=(k-1)y
Because: x > y > 0
So: 4-k2.5
Ascending power permutation and descending power permutation
First, arrange the following polynomials according to the letter X to reduce the power, and then arrange them according to the letter X to increase the power, such as 7x-5x3 power - the 5th power of 1 / 5 x + 3?
Power reduction
-1/5*x^5-5x^3+7x+3
Ascending power
3+7x-5x^3-1/5*x^5
Power reduction: - 1 / 5x ^ 5-5x ^ 3 + 7x + 3
^ 3 + 5x-5
Classmate, what's your student number? I'm the teacher of this course. Don't put your homework on the Internet.
5 times, 7x-5x3 times, + 3 times
The nonnegative integer solutions of inequality 4x-1 less than 3 are
That's 4x
Obviously 0
4x-1
Given that P is a prime number, two of the quadratic equations X & sup2; - 2px + P & sup2; - 5p-1 = 0 with respect to X are integers, so we can find the value of P
By using the root rule of Δ, we get that Δ = 4P & sup2; - 4 (P & sup2; - 5p-1) = 20p + 4
So x = [P + √ (20p + 4)] / 2 or x = [P - √ (20p + 4)] / 2
So √ (5p + 1) is an integer
Let (5p + 1) = K & sup2; K be a positive integer
So 5p = (k-1) (K + 1)
Since P is prime, 5p = 1 * 5 * P
So k-1 = 1, K + 1 = 5p or k-1 = 5, K + 1 = P or k-1 = P, K + 1 = 5
So we sort out the above three cases, omit the first one, and get k = 6, P = 7 or K = 4, P = 3
So p is 3 or 7
B ^ 2-4ac = 20p + 4 = 4 (5p + 1) 5p + 1 = a ^ 2 5p = (A-1) (a + 1) P = 3 or 7
P is a prime number. For the quadratic equation of one variable X & # 178; - 2px + P & # 178; - 5p-1 = 0 of X, both of them are integers. Find the value of P. Let 5p + 1 = K & # 178;, k > 1, 5p = (k-1) (K + 1), k-1 = 5 and P = K + 1, or K + 1 = 5 and P = k-1, P = 7 or P = 3
If the solution of the equations 4x + 3Y = 2, KX + (K-3) y = 3 satisfies x = y, the value of K is obtained
From x = y, we get: 4x + 3Y = 7Y = 2, so y = 2 / 7
KX + (K-3) y = K (x + y) - 3Y = 4K / 7-6 / 7 = 3
4k/7=27/7
k=27/4
X = y is substituted into the first formula to get x = y = 2 / 7
Substituting x = y into the second formula KX + (K-3) x = 3
2KX-3X=3
2K * 2 / 7-3 * 2 / 7 = 3, k = 27 / 4
Who can tell me about ascending power permutation and descending power permutation
seek
Ascending power arrangement: Generally speaking, it means the number of power from small to large; for example, x, the square of X, the cube of X, the fourth power of X, the eighth power of X, and the tenth power of X;
Descending power arrangement: Generally speaking, it means the number of power from large to small; for example, the 10th power of X, the 8th power of X, the 4th power of X, the cube of X, the square of X, X;
Finding the nonnegative integer solution of inequality 1-4x / 3 ≥ 1-2x + 3 / 2
1-4x/3≥1-2x+3/2
2(1-4x)≥6-3(2x+3)
2-8x≥6-6x-9
-8x+6x≥-3-2
-2x≥-5
x≤5/2
The nonnegative integer solutions of inequality 1-4x / 3 ≥ 1-2x + 3 / 2 are: 0,1,2
6-4x≥6-12x+9
x≥9/8
It is known that the two real roots of the quadratic equation x ^ 2 + MX + 1 = 0 with respect to X are p and Q. is there m such that P and Q satisfy 1 / P + 1 / Q = 1?
If it exists, find out the value of M. if it does not exist, explain the reason
Solution: there is, from the relationship between the root and coefficient of quadratic equation of one variable, P + q = - m, PQ + 1,1 / P + 1 / Q = (P + Q) / PQ = (- M) / 1 = - m, because 1 / P + 1 / Q = 1, so m = - 1
Is the above process correct? If not, write the correct process
The answer is that there is no such M
1. If the equation has a solution, the discriminant = M & # 178; - 4 * 1 ≥ 0
The solution is m ≥ 2
2. Your calculation is correct, M = - 1
It is not in the range of M
So, there is no such M
(you can substitute m for the equation, and there is no solution.)
incorrect:
x^2+mx+1=0
a=1 ,b=m c=1
b^2-4ac=m^2-4
M ^ 2-4 > 0
The solution is m > 2 or m-2
So it doesn't exist
If the two real roots of a quadratic equation with quadratic coefficient 1 are p, Q, then the equation is: (X-P) (X-Q) = 0, that is: x ^ 2 - (p q) x PQ = 0, p q (p 1) = 5, PQ (p q) = 5--
It's not correct. It doesn't consider the condition that the equation has a solution, and it should satisfy M2-4 > 0. In this way, M = - 1 is excluded, so there is no m value satisfying the meaning of the problem
If the distance between at least three points on the circle x2 + y2-4x-6y-12 = 0 and the straight line 4x-3y = m is 4, then the value range of M is ()
A. -21＜m＜19B. -21≤m≤19C. -6＜m＜5D. -6≤m≤4
Circle x2 + y2-4x-6y-12 = 0, that is, (X-2) 2 + (Y-3) 2 = 25, which means a circle with a (2,3) as its center and 5 as its radius. The distance from at least three points on the circle to the straight line 4x-3y = m is 4. The distance from the center of the circle to the straight line is less than or equal to 5-4 = 1, that is | 4 × 2 − 3 × 3 − m | 16 + 9 ≤ 1. The solution is - 6 ≤ m ≤ 4, so D
Permutation solution of descending power and ascending power
The third power of 3x - the fourth power of Y + the third power of 2x, the third power of y-xy + the second power of 4x, the second power of Y
Arrange by descending power of X ()
Arrange by ascending power of Y ()
The term that makes up the third power of polynomial X - the second power of half x + three fourths X-1 is ()
Arrange by descending power of X (3x ^ 3 + 2x ^ 3 * y + 4x ^ 2 * y ^ 2-x * y ^ 3-y ^ 4)
Arrange by ascending power of Y (- y ^ 4-x * y ^ 3 + 4x ^ 2 * y ^ 2 + 2x ^ 3 * y + 3x ^ 3)
There seems to be something wrong with your question