If (1) 2x + 5A ≤ 3 (x + 2) (2) 2 bifurcation If the system of inequalities 2x + 5A ≤ 3 (x + 2) 2 x-a < 3 x If there are solutions and each solution is not in the range of - 1 ≤ x ≤ 4, the value range of a is obtained

If (1) 2x + 5A ≤ 3 (x + 2) (2) 2 bifurcation If the system of inequalities 2x + 5A ≤ 3 (x + 2) 2 x-a < 3 x If there are solutions and each solution is not in the range of - 1 ≤ x ≤ 4, the value range of a is obtained

The solution 2x + 5A ≤ 3 (x + 2) gives x ≥ 5a-6
Solution (x-a) / 2 < x / 3 gives x < 3a
Every solution is not in the range of - 1 ≤ x ≤ 4
The solutions of two inequalities need to be satisfied
1) When 5a-62 < A3A, a > 3
5a-6 > 4, or 3a ≤ - 1 = = > a > 3
3) If 5a-6 = 3A, then a = 3 inequality has no solution
So the solution is a ≤ - 1 / 3, a > 2 and a is not equal to 3
The solution 2x + 5A ≤ 3 (x + 2) gives x ≥ 5a-6
Solution (x-a) / 2 < x / 3 gives x < 3a
And ∵ - 1 ≤ x ≤ 4
The solutions of two inequalities need to be satisfied
1）5a-6≥1
2）3a≤4
3）5a-6≤3a
By solving the three inequalities, a ≥ 1, a ≤ 4 / 3 and a ≤ 2 are obtained
The solution set of this inequality is 1 ≤ a ≤ 4 / 3
The solution 2x + 5A ≤ 3 (x + 2) gives x ≥ 5a-6
If (x-a) of 2 is less than X of 3, then x is less than 3a
From the meaning of the title, we can get - 1 ≥ 3A > 5a-6; from the solution, we can get a ≤ - 1 / 3
Or 3a ＞ 5a-6 ＞ 4, the solution is 3 ＞ a ＞ 2
That is, a ≤ - 1 / 3 or 3 ＞ a ＞ 2
If the system of inequalities 2x + 5A ≤ 3 (x + 2) x-a / 2
2x+5a≤3(x+2),x-a/2≤x/3
x≥5a-6,x≤3a/4
Because the system of inequalities has solutions
So the solution is 5a-6 ≤ x ≤ 3A / 4
So 3A / 4 ≥ 5a-6
So a ≤ 24 / 17
And every solution x is not - 1 ≤ x ≤ 4
Then 3A / 4 ＜ - 1 or 5a-6 ＞ 4
So a ＜ - 4 / 3 or a ＞ 2
So the value range of a is {a | a ＜ - 4 / 3}
As shown in the figure, in the equilateral triangle ABC, the bisectors of ∠ B and ∠ C intersect at O, and the vertical bisectors of OB and OC intersect BC at e and F. try to explore the size relationship of be, EF and FC, and explain the reasons
Conclusion: be = EF = FC (1 point) the reason is: be = EF = FC (1 point) the reason is: ∵ ABC is equilateral triangle, ∵ ABC = ∠ ACB = 60 ° (2 points), ∵ OC, OB bisection ∵ ACB, ∵ ABC, ∵ OBE = ∠ OCF = 30 ° (3 points), ∵ eg, HF vertical bisection ob, OC, ∵ OE = be, of = FC (5 points), ∵ BOE = ∠ OBE = 30 °, ∵ COF = ∠ OCF = 30 °, ∵ OEF = ∠ ofe = 60 °, ∵ triangle OEF is equilateral triangle (8 points), Of = OE = EF, be = EF = FC (10 points)
Finding the integer solution of the inequality x-3 (X-2) ≤ 8,5-half x > 2x
x-3(x-2) ≤ 8
5-x/2 > 2x
Well organized
-2x ≤ 2
5x < 10
The solution is
-1 ≤ x < 2
The integer solution is: - 1,0,1
That's right!! fantastic!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
In equilateral triangle ABC, the bisector of angle B and angle c intersects at O, and the vertical bisector of OB and OC intersects at e, F. it is proved that be = EF = FC
Connecting OE, of
Obviously OE = be (1), 0f = FC (2)
The angle OEF = 2, OBE = 60, ofe = 2, OCF = 60
So the triangle OEF is an equilateral triangle, so OE = of = ef (3)
It is proved by (1) (2) (3)
If OE and of are connected, OE = be and of = FC can be known. If OEF and ofe are 60 degrees, be = EF = FC
The solution set of inequality system 2x + 1 > 3 negative half (x-3) > 0 x + 6 < 11 is
I'll add 50 points
2X+1＞3
X>1
-1/2﹙X-3﹚＞0
X-3
3＜x＜5
As shown in the figure, in the equilateral triangle ABC, the bisectors of ∠ B and ∠ C intersect at O, and the vertical bisectors of OB and OC intersect BC at e and F. try to explore the size relationship of be, EF and FC, and explain the reasons
Conclusion: be = EF = FC (1 point) the reason is: be = EF = FC (1 point) the reason is: ∵ ABC is equilateral triangle, ∵ ABC = ∠ ACB = 60 ° (2 points), ∵ OC, OB bisection ∵ ACB, ∵ ABC, ∵ OBE = ∠ OCF = 30 ° (3 points), ∵ eg, HF vertical bisection ob, OC, ∵ OE = be, of = FC (5 points), ∵ BOE = ∠ OBE = 30 °, ∵ COF = ∠ OCF = 30 °, ∵ OEF = ∠ ofe = 60 °, ∵ triangle OEF is equilateral triangle (8 points), Of = OE = EF, be = EF = FC (10 points)
Find the integer solution of the system of inequalities {x-3 (2-x) > - 8, half X - (2x-3) ≥ quarter
1、x-6+3x>-8
x>-1/2
2、2x-8x+12≥1
11/6≥x
To sum up, it is 0,1
The first formula gives x > - 1 / 2
The second formula gives X
Triangle ABC is equilateral triangle, BD is the middle line of AC side, extend BC to point E, make CE = CD, prove: point D is on the vertical bisector of line be
Make DF ⊥ be, the perpendicular foot is f, because the triangle ABC is equilateral triangle, so ⊥ ABC = ⊥ BCD = 60 ° because CD = CE, so ⊥ e = ⊥ CDE, and ⊥ BCD = ⊥ e + ⊥ CDE = 60 ° so ⊥ e = ⊥ BCD / 2 = 30 ° because BD is the middle line of AC side, and the triangle ABC is equilateral triangle, so BD bisects ⊥ ABC, so ⊥ CBD = 30 ° so
Syndrome: make be vertical line through D point and cross be at f point
A triangle ABC is an equilateral triangle
∴∠C=60º，∠FDC=30º
∴FC=½CD=¼BC
Also, CE = CD = &? 189; BC
∴BE=1.5BC，BF=BC-CF=¾BC=½BE
The DF is vertically bisected in be.
Inequality KX ^ 2-2x + 1-k
One person does this: KX ^ 2-2x + 1-k = K (x ^ 2-1) - 2x + 1
Here we take K as an independent variable. - 2 ≤ K ≤ 2 is the domain of function
The question is: what condition does x satisfy in this domain