Inequality 3 (x + 1) > = 5x-2, then / 2x-5/= The title is / 2x-5 / =? I only know that / 2x-5 / is less than or equal to 0 I don't know what to fill in

Inequality 3 (x + 1) > = 5x-2, then / 2x-5/= The title is / 2x-5 / =? I only know that / 2x-5 / is less than or equal to 0 I don't know what to fill in

3(x+1)>=5x-2,
3x+3>=5x-2,
Then we know that the absolute value of negative number is equal to its opposite number, namely 5-2x
2x=5x-2
x=5x-2
So: 5 > = 2x;
So 5-2x > = 0;
Then / 2x-5 / = 5-2x
Extrapolate back
3(x+1)>=5x-2
3X+3=5x-2
So the shift is 2x-5 = 0
The absolute value of 0 or 0
The absolute value of 2x ≤ 5 x ≤ 5 / 2 2x-5 is 0
From the inequality, we can know 3x > = 5x-5, that is, we can know 0 > = 2x-5, so I really don't know what the following is equal to, because there may be too many numbers
2X
Solving inequalities, 7x-10
7x-10
Solving inequality x ^ 3-5X ^ 2 + 2x + 8
x³-2x²-(3x²-2x-8)
Polynomial an * x ^ n + +A1 * x + A0 = 0, the possible rational root is the divisor of AO / an,
The possible roots in this question are ± 1, ± 2, ± 4, ± 8
Substituting test, we get (X-2) (x-4) (x + 1)
The minimum negative integer solution satisfying inequality 3 (1 + Half x) is - 0.5x-7
3（1+x/2）≥-0.5x-7;
3+1.5x≥-0.5x-7;
2x≥-10;
x≥-5;
The minimum integer solution of Z is - 5;
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The known set a = {X / 2 ≤ x ≤ 10}, B = {X / log2 (x-1)
log2(x-1)
If the absolute value of a is equal to negative a, then the value range of a is (); if the absolute value of X is greater than x, then the value range of X is (), which requires steps
1. If | a | = - A, then - a > = 0, so ax
If x > = 0, then | x | = x, does not conform to the meaning of the title
If XX, then | x | = - x > X
So x
1. If | a | = - A, then - a > = 0, so ax
If x > = 0, then | x | = x, does not conform to the meaning of the title
If XX, then | x | = - x > X
So x
All formulas of Physics
High school physics part: 1. Motion of particle (1) ---- linear motion 1) uniform speed linear motion 1. Average velocity VT = s / T (definition) 2. Useful inference vt & sup2; - vo & sup2; = 2As 3. Middle time velocity VT / 2 = VT + VO / 2 4. Final velocity VT = VO + at 5. Middle position velocity vs / 2
(3x + 2Y) / 4 = (2x + y + 2) / 5 = - (x + 5Y) / 3 by addition and subtraction elimination method
(3x+2y)/4=(2x+y+2)/5=-(x+5y)/3
Break it down into two equations
(3x+2y)/4=(2x+y+2)/5………… （1）
(3x+2y)/4=-(x+5y)/3………… （2）
The results are as follows
5(3x+2y)=4(2x+y+2)
15x+10y=8x+4y+8
7x+6y=8………… （3）
The results are as follows
3(3x+2y)=-4(x+5y)
9x+6y=-4x-20y
13x=-26y
x=-2y………… (4) (copy it, please)
It can be obtained from formula 4
7x+14y=0 .(5)
Five minus three
8y=-8
We get y = - 1
Then we get x = 2
① 2X + Y-Z = 0, ② x-3y-6z = 5, ③ 3x + 2Y = 0, ① × 6 leads to 12x + 6y-6z = 0, ④ - ② leads to 11x + 9y = - 5, ③ × 4.5 leads to 13.5x + 9y = 0, ⑥ - ⑤ leads to 2.5x = - 5?
It is known that the polynomial a (the cube of X - the square of X + 3x) + B (the square of 3x + 4x) + the cube of 2x - 8 is a quadratic trinomial, and the value is 3 when x = - 1
When x = - 2, the value of the polynomial
A (x ^ 3-x & # 178; + 3x) + B (3x & # 178; + 4x) + 2x ^ 3-8 = (a + 2) x ^ 3 + (3b-a) x & # 178; + (3a + 4b) X-8 is a quadratic trinomial, which shows that a + 2 = 0, a = - 2 becomes (3b + 2) x & # 178; + (4b-6) X-8, when a = - 1, the former = (3b + 2) - (4b-6) - 8 = - B = 3B = - 3, the former becomes - 7x & # 178; - 18x-8a = - 2
Solution: a (x ^ 3-x ^ 2 + 3x) + B (3x ^ 2 + 4x) + 2x ^ 3-8 = ax ^ 3-ax ^ 2 + 3ax + 3bx ^ 2 + 4bx + 2x ^ 3-8
=(a+2)x^3+(3b-a)x^2+(4b+3a)x-8.
A + 2 = 0, a = - 2
The original polynomial = (3b + 2) x ^ 2 + (4b-6) X-8
When x = - 1, the value of polynomial is 3
Solution: a (x ^ 3-x ^ 2 + 3x) + B (3x ^ 2 + 4x) + 2x ^ 3-8 = ax ^ 3-ax ^ 2 + 3ax + 3bx ^ 2 + 4bx + 2x ^ 3-8
=(a+2)x^3+(3b-a)x^2+(4b+3a)x-8.
A + 2 = 0, a = - 2
The original polynomial = (3b + 2) x ^ 2 + (4b-6) X-8
When x = - 1, the value of polynomial is 3
(3b+2)*(-1)^2+(4b-6)*(-1)-8=3.
3b+2-4b+6-8=3.
-b=3.
∴b=-3.
When x = - 2, the original polynomial = [3 * (- 3) + 2} * (- 2) ^ 2 + [4 * (- 3) - 6] * (- 2) - 8
=(-7)*4+[-18*(-2)]-8.
=-28+36-8.
=0
When x = - 2, the value of the polynomial is 0
Given the set a = {x | 3 ≤ x ＜ 6}, B = {x | 2 ＜ x ＜ 9}. Find Cr (a ∩ b), (CRB) ∪ a
∩ a ∩ B = {x | 3 ≤ x ＜ 6} (2 points) ∩ Cr (a ∩ b) = [x | x ＜ 3 or X ≥ 6} (4 points) ∩ CRB = {x | x ≤ 2 or X ≥ 9} (6 points) ∪ a = {x | x ≤ 2 or 3 ≤ x ＜ 6 or X ≥ 9} (8 points)