1. Negative integers with absolute values greater than 1 and less than 3 are____ ; the largest negative integer is_____ The number with the smallest absolute value is______ 2. If | - a | - a = 0, then a is_____ Ab | ab | = - 1 (b)

1. Negative integers with absolute values greater than 1 and less than 3 are____ ; the largest negative integer is_____ The number with the smallest absolute value is______ 2. If | - a | - a = 0, then a is_____ Ab | ab | = - 1 (b)

1、-2 -1 0
2. Nonnegative positive

First aid of multiplication and division of rational number in mathematics of grade one!
|A-2|+|-B+6|+|C-3|=0

|A-2|+|-B+6|+|C-3|=0
The absolute value is a non negative number, and those with absolute values are all 0
So, a = 2, B = 6, C = 3

If | x | + | y | = 0, then x=__________ ,y=__________ .
How to answer

∣ x ∣ ≥ 0, ∣ y ∣ ≥ 0, and | x | + | y | = 0,
∴∣x∣=0,∣y∣=0
∴x=0,y=0

When a student solves the equations ax + by = 1,2ax + 3By = - 1, he misinterprets the second y symbol to get x = 2, y = 1, and calculates the values of a and B

x=2,y=1
So 2A + B = 1 (1)
4a-3b=-1 (2)
(1)×3+(2)
6a+4a=3-1
10a=2
therefore
a=1/5
b=1-2a=3/5

1640 + 3636 △ 18 × 2.4 and 9.728 △ 3.2 △ 5 should be calculated by simple method
1640 + 3636 △ 18 × 2.4 and 9.728 △ 3.2 △ 5 should be calculated by simple method

1) Original formula = 1640 + (3600 + 36) △ 18 × 2.4
= 1640+（200+2）×2.4
= 1640+480+4.8
= 2124.8
2）9.728÷3.2÷5=9.728÷（4x0.8x5）
=9.728÷16
=0.608

Proposition p: the first-order function y = (A-1) x + 2 is a decreasing function on R; proposition q: the solution set of the inequality AX2 < AX-1 about X is Ø (1) if proposition q is a true proposition, try to find the value range of a; (2) if "P and Q" are true propositions, try to find the value range of a; (3) if "P or Q" are true propositions, try to find the value range of A

∵ the linear function y = (A-1) x + 2 is a decreasing function on R ∵ A-1 < 0 That is to say, the solution set of the inequality AX2 ＜ AX-1 of P: a ＜ 1 ∵ about X is Ø. AX2 ax + 1 ≥ 0 holds (I) when a = 0, 1 ≥ 0 holds, which is consistent with the meaning of the problem. (II) when a ≠ 0, a ＞ 0 △ = A2 − 4A ≤ 0 solution can be obtained, 0 ＜ a ≤ 4. In conclusion, 0 ≤ a ≤ 4 is Q: 0 ≤ a ≤ 4 (1) if proposition q is true proposition, then 0 ≤ a ≤ 4 (2) if "P and Q" are true propositions, then propositions P and Q are true propositions (3) if "P or Q" is true, then at least one of P and Q is true. If P and Q are false, then a ≥ 1A < 0 or a > 4, that is, a > 4. If P or q is true, then a ≤ 4

The method of 17 out of 12 and 7 out of 18 Division

The answer to your question is as follows: first, find out the least common multiple of 12 and 18, and use the short division method to calculate that their least common multiple is 36, then their denominator is 36, the denominator expands, the numerator also expands, the size of the fraction remains unchanged, 12 expands 3 times, the numerator 17 also expands 3 times

Inequality 1 / 2x + 3

If the inequality holds, 1 / 2x + 3 is less than zero, that is, y is less than 0, so the value of function y is less than 0, so the image should be above the x-axis

Simple calculation: 27 out of 29 times 28 plus 27 out of 29

27

If the image of a quadratic function is known to pass through points (4, - 3) and x = 2, the function has a maximum value of - 1, the analytic expression of the quadratic function is obtained

Let the vertex formula y = a (x-k) ^ 2 + H, because when x = 2, there is a maximum value of - 1, so k = 2, H = - 1, it becomes y = a (X-2) ^ 2-1, and then (4, - 3) is brought in to solve a