The number of nonnegative integer solutions of inequality system 3 − 2x > 02x − 7 ≤ 4x + 7 is______ One

The number of nonnegative integer solutions of inequality system 3 − 2x > 02x − 7 ≤ 4x + 7 is______ One

3 − 2x ＞ 0 (1) 2x − 7 ≤ 4x + 7 (2) from (1), we get x ＜ 32; from (2), we get x ≥ - 7. Then its solution set is - 7 ≤ x ＜ 32
8. It is known that the two real number roots of the equation x2 + PX + q = 0 are respectively 1 smaller than the two real number roots of the equation x2 + QX + P = 0. The solution is a quadratic equation with two variables
The conclusion is not complete
Let two real number roots of x2 + PX + q = 0 be X1 and X2 respectively, then the two real number roots of x2 + QX + P = 0 are x1-1 and x2-1. According to Weida's theorem, there are X1 + x2 = - P (1), x1x2 = q (2). X1-1 + x2-1 = - Q (3), (x1-1) (x2-1) = P (4). In this way, - P-2 = - Q can be obtained from (1) (3), and Q + P + 1 = P can be obtained from (2) (4) (1), then the values of P and Q can be obtained, and the equation can be constructed
If there are two tangents passing through point P (2,1) to make circle C: x2 + y2-ax + 2ay + 2A + 1 = 0, then the value range of a is ()
A. A ＞ - 3B. A ＜ - 3C. - 3 ＜ a ＜− 25d. - 3 ＜ a ＜− 25 or a ＞ 2
From the meaning of the title, we can see that d = - A, e = 2A, f = 2A + 1, so D2 + e2-4f = A2 + (2a) 2-4 (2a + 1) > 0, simplify 5a2-8a-4 > 0, that is (5a + 2) (A-2) > 0, solve a > 2 or a < - 25 ①; and point P is substituted into the equation of circle, get 22 + 12-2a + 2A + 2A + 1 > 0, solve a > - 3 ②, then the value range of a is: - 3 < a < - 25 or a > 2, so choose D
Decreasing power permutation polynomials of integers
1) If the fourth power of polynomial 5x, the A-1 power of y-2x, the third power of Y + the B + 1 power of 6x, the fourth power of Y - the second power of 3xy are arranged according to the descending power of X, try to find the value of b-a
2) The quadratic power of polynomial 5x, the cubic power of Y + y and the cubic power of - 3xy-x are arranged according to the ascending power of the letter y
（1）
a-1=3
b+1=2
So a = 4, B = 1
So B-A = - 3
（2）
-x+(5x^2-3x)y+y^3
Finding positive integer solutions of inequality (2-3x) / 4 - (X-5) > (- 4x + 1) / 6 + 2 / 3
Such as the title
The two sides are multiplied by 12 at the same time
21-12x>20-8x
Then - 4x > 8
So x > - 2
The minimum positive integer solution is - 1
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A quadratic equation with one variable: x ^ 2 + PX + q = 0 has two real roots with different sign if and only if Q < 0
correct
When Q0, there must be real roots, and X1 * x2 = Q
There are two different signs
If there are two different signs, then X1 * x2 = Q
Let m be the upper moving point of circle x2 + y2-6x-8y = 0, o be the origin, and n be the point of ray OM. If | om | · | on | = 120, the trajectory equation of point n is obtained
Let the coordinates of M and n be (x1, Y1), (x, y) respectively. Let | om | · | on | = 120, then we get X21 + y21 · x2 + y2 = 120. When x1 ≠ 0, X ≠ 0, there is YX = y1x1. If YX = y1x1 = k, there is y = KX, Y1 = kx1, then the original equation is X12 + k2x12-6x1-8kx1 = 0
Place a polynomial in descending order of a letter. Where is the constant term?
The polynomials XY & sup3; - 8x & sup2; Y-X & sup3; Y & sup2; - 6 are arranged in descending order of Y,
Put it last
The permutation result is XY & sup3; - X & sup3; Y & sup2; - 8x & sup2; y-6 (the powers of Y are 3,2,1,0 in turn)
The integer solution of inequality 6x + 15 > 2 (4x + 3), 2x-1 / 3 ≤ 1 / 2x-2 / 5
6x+15＞8x+6;
2x＜9;
x＜9/2;
2x-1/3≤x/2-2/5;
3x/2≤11/15;
x≤22/45;
∴x≤22/45;
There are countless integer solutions
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To prove that the quadratic equation x * 2 + PX + q = 0 with real coefficients has two different sign roots if and only if Q < 0
If there are two different sign roots
Then x1x2