The known set a = {X - 1}

The known set a = {X - 1}

B = {y, y = x + A, X ∈ a} = (- 1 + A, 2 + a),
C = {Z Z = 2x + 3, X ∈ a} = (1,7),
B = B ∩ C, B is a subset of C,
-1+a>=1,2+a
By - 1
If the value of the algebraic formula 2-x / 4 - x is not less than - 3, then the value range of X is
Quarter (2-x) - x
(4-x)/(2-x)≥-3
∴(4-x)/(2-x)+3≥0
∴(10-4x)/(2-x)≥0
∴(2x-5)/(x-2)≥0
Ψ x ≥ 5 / 2 or x < 2
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(4-x)/(2-x)≥-3
-3(2-x)≤4-x
3x+x≤4+6
x≤10÷4=2.5
(2-x-4) - x is greater than or equal to - 3
2-5x greater than or equal to - 12
5x less than or equal to 14
X is less than or equal to 2.8
When x is less than 2 or X is not less than 2.5, this fraction holds
When x > 2, 4-x = 2.5
When x = - 3 (2-x), i.e. x = 2.5 or X
How about 35 (X-2) - 15 (5x-6) = (22x-63) - 21 (3x-4)?
35(x-2)-15(5x-6)=(22x-63)-21(3x-4)
35x-70-75x+90=22x-63-63x+84
-45x+20=-41x+21
45x-41x=20-21
4x=-1
x=-1/4
35X-70-75X+90=22X-63-63X+84
The next step is to solve the equation.
35X-70-75X+90=22X-63-63X+84
The solution is x = 1
Don't cheat children on the first floor
Square difference formula 2010 × 2012 =? 100 & # 178; - 99 & # 178; + 98 & # 178; - 97 & # 178; +... + 2 & # 178; - 1
2010*2012=（2011-1）*（2011+1）=2011²- 1²100²-99²+98²-97²+...+2²-1
=(100-99)(100+99)+(98-97)(98+97)+…… +(2-1)(2+1)
=100+99+98+97+…… +2+1
=(1+100)*100/2
=5050
If the circle (x-a) ^ 2 x y ^ 2 = 9 and the ellipse x ^ 2 / 9 x y ^ 2 / 4 = 1 have a common point, then the value range of the real number a
It can be seen from the figure that when the abscissa of the center of the circle is - 6, the rightmost point of the circle is (- 3,0), tangent to the ellipse;
When the abscissa of the center is 6, the leftmost point of the circle is (3,0), which is tangent to the ellipse;
So, - 6 ≤ a ≤ 6
Given that the value of X-5 + 1 is not less than that of X + 1-1, we can find the value range of X
（2X-10）/6 +1≥(3X+3)/6 -1
(X+13)/6≤2
X≤-1
35（x-2）-15（5x-6）=22x-63-21（3x-4）
35（x-2）-15（5x-6）=22x-63-21（3x-4），           35x-70-75x+90=22x-63-63x+84，         35x-75x-22x+63x=...
It is known that the quadratic function y = 1 / 2x & # 178; + X-5 / 2
① Find the vertex coordinates and the axis of symmetry. 2. If the two intersections of the quadratic function image and the X axis are AB, find the length of ab
y=1/2x²+x-5/2
=(x+1)²/2-3
So the vertex coordinates are (- 1,3) and the axis of symmetry is x = - 2
Let y = 0
Then (x + 1) & # / 2-3 = 0
(x+1)²=6
x+1=±√6
x=-1±√6
So | ab | = (- 1 + √ 6) - (- 1 - √ 6) = 2 √ 6
If you don't understand, I wish you a happy study!
Given that the ellipse x ∧ 2 / 9 + y ∧ 2 / 4 = 1 and the circle (x-a) ∧ 2 + y ∧ 2 = 9 have a common point, then the value range of a?
If you don't use the root graph, how can you do it quickly?
When a = 0, there are just two common points between the circle and the ellipse. The center of the circle should not move left or right more than 6 units, otherwise the ellipse and the circle have no common point
-6≤a≤6.
Or jointly:
X Λ 2 / 9 + y Λ 2 / 4 = 1, y ^ 2 = 4 (1-x ^ 2 / 9)
（x-a）∧2+y∧2=9
x^2-2ax+a^2+4-4x^2/9=9
5x^2/9-2ax+a^2-5=0
Δ≥0
-6≤a≤6.
[welcome to ask, thank you for adopting!]
When the value of X is in what range, the value of X-1 / 3 + X-1 / 2 is not less than that of X-1 / 6
(x + 1) / 3 - (x-1) / 2 ≥ (x-1) / 6
2（X+1）-3（X-1）≥X-1,
So, X ≤ 3