Given the set a = {1,3, - A ^ Z}, B = {1, a + 2}, is there a real number a such that B ≤ a?

Given the set a = {1,3, - A ^ Z}, B = {1, a + 2}, is there a real number a such that B ≤ a?

a+2=3
A=1
sure
therefore
There exists a = 1 such that B is contained in a
a+2=3
A=1
sure
therefore
There exists a = 1 such that B is contained in a
When x, the value of the algebraic formula 2x + 5 is not greater than zero
What's after x
X is less than or equal to minus five-thirds
2x+5=
It is known that the function f (x) defined on (- 1,1) is a decreasing function, and f (A-1) > F (2a)
f(a-1)>f(2a)
Decreasing function
A-1
-1
X & # 178; + Y & # 178; = 3 x & # 178; - Y & # 178; = 0
x²+y²=3 x²-y²=0
Add to get
2x^2=3
x=±√6/2
Subtraction
2y^2=3
y=±√6/2
So there are four sets of solutions
x²+y²+（x²-y²）=2x^2=3
x=±√6/2
x²+y²-（x²-y²）=2y^2
y=±√6/2
x²+y²=3 ①
x²-y²=0 ②
①+②
2x²=3
x=±√6/2
①-②
2y²=3
y=±√6/2
Add two formulas, 2x ^ 2 = 3
So the square of x equals the square of Y equals three-thirds.
So x = y = three-thirds of the positive and negative radicals
To simplify, it's equal to plus or minus half root six.
Given the set a = {x | - 1 ≤ x0}, satisfying B ∪ C = C, find the value range of real number a
(1) It can be seen from the meaning that B = {x is greater than or equal to 2}, so the intersection of two sets is {2 is less than or equal to X is less than 3}
(2) It is easy to get C = {x is greater than - A / 2}, and B is a subset of C, so - A / 2 is less than 2, so a is greater than - 4
When x takes what value, the value of 2x + 8 is (1) positive; (2) negative; (3) zero; (4) no more than 4
(1)(2x+8)/3>0,2x+8>0,x>-4;
(2)(2x+8)/3
Solving the equation 3x ^ 2 + 5x-1 = 0
Because △ = 5 & # 178; - 4 × 3 × (- 1) = 37 ＞ 0
So x = (- 5 ± √ 34) / (2 × 3)
x=（-5±√37）/6
Headache? Ask: you solve it
(2x + 1) (4x & # 178; - 2x + 1) - 2x (4x & # 178; + 1) = 0,
（2x+1）(4x²-2x+1)-2x(4x²+1）=0
8x³+1-8x³-2x=0
2x=1
x=1/2
Let a = {x | X-2 / 2x + 1 > 1}, B = {x | x-a}|
A:(x-2-2x-1)/(2x+1)＞0;
(-x-3)/(2x+1)＞0;
(x+3)/(2x+1)＜0;
-3＜x＜-1/2;
B:-2＜x-a＜2;
-2+a＜x＜2+a;
∩ a ∩ B = empty set
2 + a ≤ - 3 or - 2 + a ≥ - 1 / 2;
A ≤ - 5 or a ≥ 3 / 2;
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Find the value range of the algebraic formula (X & # 178; - 2x-3) / (2x & # 178; + 2x + 1)
y=(X²-2X-3)/(2X²+2X+1)
Discriminant analysis
2x^2y+2xy+y-x^2+2x-3=0
x^2(2y-1)+x(2y+2)+y-2=0
Because x belongs to any number, y holds, so the equation has root discriminant 0
4（y+1)^2-4(2y-1)(y-2)》0
(7-3√5)/2
-4 ~ ~ 0.5, the minimum value of the upper part is - 4, the minimum value of the lower part is 1, and the maximum value of the limit existence is 0.5