If there are only 1,2,3 integer solutions for the system of X inequalities (5x-a ≥ 0,4x-b ≤ 0), then there are several ordered pairs (a, b) composed of integers a and B which are suitable for the system of X inequalities

If there are only 1,2,3 integer solutions for the system of X inequalities (5x-a ≥ 0,4x-b ≤ 0), then there are several ordered pairs (a, b) composed of integers a and B which are suitable for the system of X inequalities

The solution of the system of X inequalities (5x-a ≥ 0, 4x-b ≤ 0) is
a/5≤x≤b/4
The integer solution is only 1,2,3
So there are
Zero
A mathematics problem of grade three in junior high school (quadratic equation with one variable)
x²+ax=(x+1/2)²+b
A.a=1,b=1/4
B.a=1 b=-1/4
C.a=0 b=-1/2
D.a=2 b=1/2
solution
x²+ax=(x+1/2)²+b
=x²+x+1/4+b
∴a=1
b+1/4=0
∴a=1,b=-1/4
Choose B
B.a=1 b=-1/4
x²+ax=(x+1/2)²+b
=x²+x+1/4+b
a=1,1/4+b=0
Then, a = 1, B = - 1 / 4
Given that real numbers x and y satisfy the following conditions: 1 is less than or equal to x, 4 is less than or equal to x, 3 is less than or equal to y, 6 is less than or equal to y, then the value range of x-2y is?
Negative 11 less than or equal to x-2y less than or equal to negative 2
Directly.
1≤x≤4， 3≤y≤6
Multiply y by 2 to get 6 ≤ 2Y ≤ 12
x-2y -5≤x-2y≤-6
Adopt!!!
Just bring it in! It's simple
Polynomials 3A & sup2; B-A & sup3; B & sup3; - 1-ab & sup2; are arranged in ascending order of letter A and descending order of letter B
The polynomials 3A & sup2; B-A & sup3; B & sup3; - 1-ab & sup2; are arranged according to the ascending power of A? In descending order of the letter B is?
The ascending order of a is - A & sup3; B & sup3; + 3A & sup2; b-ab & sup2; - 1
The descending order of B is - 1 + 3A & sup2; b-ab & sup2; - A & sup3; B & sup3;
The ascending index is arranged from large to small, the descending index is arranged from small to large, and the index without unknowns (such as: - 1) is 0
The ascending order of a is - A & sup3; B & sup3; + 3A & sup2; b-ab & sup2; - 1
The descending order of B is - 1 + 3A & sup2; b-ab & sup2; - A & sup3; B & sup3;
If inequality 4x-a
Let me help you answer: first of all, your answer is right. Because there must be 1 and 2 in the solution. So when ＜ 3, we should also guarantee the integer 2, so a / 4 must be > = 2
Let me answer your question: for example, if a is 8, then X
8 ≤ a ＜ 12, 2 ≤ A / 4 ＜ 3, that is, the minimum range of X ≤ A / 4 is x ≤ 2, the maximum is x ＜ 3, and the positive integer taken by X is 1
How can there be no integer solution of x = 1?
When a = 5 is 4 * 1-5 = - 1 < 0,
When 8 ≤ a < 12, it is rare for a = 5 to have integer solution 1
When a = 8,4x-8
Quadratic equation of one variable-
If C (C is not equal to 0) is the root of the quadratic equation x & # 178; + BX + C = 0, then the value of C + B is ()
A.1 B.-1 C.2 D.-2
Because C is the root of the equation x & # 178; + BX + C = 0,
Then, C & # 178; + BC + C = 0
And because C ≠ 0, both sides of the above formula divide by C,
C + B + 1 = 0
There is C + B = - 1
Therefore, this question is B. - 1
If C (C is not equal to 0) is the root of the quadratic equation x & # 178; + BX + C = 0 with respect to X
Let the other root be x2
cx2=c
x2=1
Substituting into the original equation
1+b+c=0
b+c=-1
Because C is the root of the quadratic equation x & # 178; + BX + C = 0 with respect to x, we substitute C into the equation c & # 178; + BC + C = 0
Because C is not equal to 0 and C is reduced, C + B + 1 = 0, that is, C + B = - 1
Choose B
C is the root of the equation x & # 178; + BX + C = 0, then:
c²＋bc＋c=0
Because C is not equal to 0, then:
c＋b＋1=0
b＋c=－1
Select [b]
Given that the real number x is less than or equal to y, less than or equal to Z, and satisfies the following conditions: 3Y + 2Z = 3 + X, 3Y + Z = 4-3x, (1) find the value range of X; (2) find the maximum and minimum value of X + y + Z
3Y+2Z=3+X.(1)
3Y+Z=4-3X.(2)
(1)-(2):z=4x-1
2*(2)-(1):3y=5-7x
X ≤ y --- > 3x ≤ 5-7x, 10x ≤ 5, X ≤ 1 / 2
Y ≤ Z --- > 5-7x ≤ 3 (4x-1), 19x ≥ 8, X ≥ 8 / 19
The value range of X is 8 / 19 ≤ x ≤ 1 / 2
2)
x+y+z=x+(5-7x)/3+4x-1=(8x+2)/3
When x = 1 / 2, the maximum value = (4 + 2) / 3 = 2
When x = 8 / 19, the minimum value = (8 * 8 / 19 + 2) / 3 = 34 / 19
(1) Formula: 3Y + 2Z = 3 + X
(2) Formula: 3Y + Z = 4-3x
(1) Formula - (2): z = 4x-1
Let x = 1 / 3
(1) Formula - 2 * (2) after formula y = 5 / 3-7 / 3x
Put x
If x + 2 is a factor of polynomial X3 + x2 + ax + B and 2A2 + 3AB + B2 ≠ 0, then the value of fraction AB2 − 4A3 + B3 − 4a2b2a2 + 3AB + B2 is______ .
∵ x + 2 is a factor of the polynomial X3 + x2 + ax + B. when x = - 2, the polynomial is equal to 0, and the solution is b-2a = 4. The original formula = (a + b) (B2 − 4a2) (2a + b) (a + b) = (B − 2a) (B + 2a) 2A + B = b-2a = 4
Known inequality 4x-a about X
4x-a
② It's better to say why
I know that ax 2 + BX + C = 0 can also be written as x 2 + (x1 + x2) x + x1x2 = 0
x1+x2=-b/a x1x2=c/a...
So don't worry about it
Given 2x ② - 19x + M = 0, one of which is 1, find the other and the value of M
Given that x (2) - 2x-1 = 0, a quadratic equation with one variable is obtained by using the relationship between the root and the coefficient so that the root is the square of the root of the original equation
1. X is another root
2x ② - 19x + M = 0, one of which is 1
1+x=19/2
x=17/2
1*x=m/2
Then M = 17
2.x②-2x-1=0
x1+x2=2
x1x2=-1
The root of the equation is X1 ^ 2, X2 ^ 2
x1^2+x2^2=(x1+x2)^2-2x1x2=4+2=6
x1^2*x2^2=(x1x2)^2=1
So the quadratic equation of one variable is x ^ 2-6x + 1 = 0
(1)
Substituting x = 1 into the original equation, we get m = 17
Substituting M = 17 into the solution, we get X1 = 1, X2 = 17 / 2
(2) I don't understand. Are two roots the square of their two roots?