As shown in the figure, in △ ABC, ad bisects ∠ BAC, de ⊥ AB, DF ⊥ AC, the perpendicular feet are e and f respectively, and BD = CD

As shown in the figure, in △ ABC, ad bisects ∠ BAC, de ⊥ AB, DF ⊥ AC, the perpendicular feet are e and f respectively, and BD = CD

It is proved that ∵ ad bisects ⊥ BAC, de ⊥ AB, DF ⊥ AC, ∵ ead = ≁ fad, ≁ AED = ≌ AFD = 90 °. ∵ ad = ad, ≌ AED ≌ AFD. ≌ AE = AF, de = DF. ≁ BD = CD, ≌ bed ≌ CFD (HL). ≌ be = CF
It is known that X and y satisfy the system of inequalities: Y > = 1; X + y = 0; and 0
In the plane rectangular coordinate system (about X, y), make three straight lines: y = 1, y = 3-x, y = x + 1
The triangle region (including boundary) surrounded by these three lines is the solution set of inequality system: {Y > = 1; X + y = 0;}
The three vertices of the triangle region are (0, 1), (1, 2), (2, 1)
When x and Y fall at the three vertices, respectively
X = 0, y = 1, then ax + by = B, that is, 0 ≤ B ≤ 2
X = 1, y = 2, then ax + by = a + 2B, that is, 0 ≤ a + 2B ≤ 2
X = 2, y = 1, then ax + by = 2A + B, that is, 0 ≤ 2A + B ≤ 2
It can be checked that for the case that X and Y fall on the points of the triangle region (including the boundary) except for the three vertices, the value range of a and B determined must contain the solution set of the above inequality system about a and B, that is:
All the inequalities 0 ≤ ax + by ≤ 2 about a and B established by the value combination (x, y) of the above inequality system about X and y, their intersection is the solution set of the inequality system about a and B {0 ≤ B ≤ 2; 0 ≤ a + 2B ≤ 2; 0 ≤ 2A + B ≤ 2}
The plane rectangular coordinate system (about a and b) is also used to solve the above inequality system {0 ≤ B ≤ 2; 0 ≤ a + 2B ≤ 2; 0 ≤ 2A + B ≤ 2}. It can be seen that the solution set is a trapezoidal region, and the four vertex coordinates are (0, 0), (- 2 / 3, 4 / 3), (2 / 3, 2 / 3), (1, 0) respectively
Then (B + 2) / (a + 1) is obtained for the four vertices
A = 0, B = 0, then (B + 2) / (a + 1) = 2
A = - 2 / 3, B = 4 / 3, then (B + 2) / (a + 1) = 10
A = 2 / 3, B = 2 / 3, then (B + 2) / (a + 1) = 8 / 5
A = 1, B = 0, then (B + 2) / (a + 1) = 1
It can be checked that for the case that a and B fall on the points other than the four vertices in the trapezoid region, (B + 2) / (a + 1) must be between the minimum value 1 and the maximum value 10 of the above four values
So the value range of (B + 2) / (a + 1) is [1,10]
1 or 10
There are two points m.n.d on the straight line of AB AC on both sides of equilateral triangle ABC, which are the outer point of triangle ABC, and the angle MDN = 60 and the angle BDC = 120
BD = DC when m.n moves on the line ab. AC, BM.NC.MN The quantitative relationship between them
(1) In this case, Q L = 2 / 3 (2) conjecture: the conclusion is still valid. It is proved that as shown in the figure, extend AC to e, make CE = BM, connect de. ∵ BD = CD, and ∵ BDC = 120 ° and ∵ DBC = ∵ DCB = 30 °. ABC is an equilateral triangle and ∵ MBD = ∵ NCD =
It is known that the solution set of inequality ax2-3x + 6 ＞ 4 is {x | x ＜ 1 or X ＞ B}, (1) finding a, B; (2) solving inequality AX2 - (AC + b) x + BC ＜ 0
(1) Because the solution set of the inequality ax2-3x + 6 ＞ 4 is {x | x ＜ 1 or X ＞ B}, X1 = 1 and X2 = B are the two real roots of the equation ax2-3x + 2 = 0, and B ＞ 1. From the relationship between the roots and the system, we get 1 + B = 3A1 × B = 2A, and the solution a = 1b = 2, so we get a = 1b = 2 B = 2, so the inequality AX2 - (AC + b) x + BC < 0, that is, X2 - (2 + C) x + 2C < 0, that is, (X-2) (x-C) < 0. ① when C > 2, the solution set of inequality (X-2) (x-C) < 0 is {x | 2 ＜ x ＜ C}; ② when C < 2, the solution set of inequality (X-2) (x-C) < 0 is {x | C ＜ x ＜ 2}; ③ when C = 2, the solution set of inequality (X-2) (x-C) ＜ 0 is ∈. In conclusion: when C > 2, the inequality (X-2) (x-c-c-c-c-c) does not exist The solution set of equation AX2 - (AC + b) x + BC ＜ 0 is {x | 2 ＜ x ＜ C}; when C ＜ 2, the solution set of inequality AX2 - (AC + b) x + BC ＜ 0 is {x | C ＜ x ＜ 2}; when C = 2, the solution set of inequality AX2 - (AC + b) x + BC ＜ 0 is ∈
There are two points m and N on the straight line AB and AC on both sides of equilateral △ ABC. D is a point outside △ ABC, and ∠ MDN = 60 °, BDC = 120 ° BD = DC. Explore: when m and N move on the straight line AB and AC respectively, the quantitative relationship between BM, NC and Mn, and the relationship between the perimeter Q of △ amn and the perimeter l of equilateral △ ABC. (1) as shown in Figure 1, when m and N are on AB and AC, and DM = DN, the relationship between BM, NC and Mn is as follows What is the quantitative relationship between them___ In this case, QL = 0___ (2) as shown in Fig. 2, are the two conclusions of conjecture (1) still valid when DM ≠ DN on points m, n edges AB and AC? Write out your conjecture and prove it. (3) as shown in Figure 3, when m and N are on the extension line of edge AB and Ca respectively, if an = x, then q = 0___ (denoted by X, l)
(1) In this case, QL = 23. (2) conjecture: the conclusion is still valid. Prove: as shown in the figure, extend AC to e, make CE = BM, connect de. ∵ BD = CD, and ∵ BDC = 120 °, and ∵ DBC = ∵ DCB = 30 °, and ∵ ABC is an equilateral triangle, ∵ MBD = ∵ NC
If the positive integer solution of the system of inequalities 2x + 38 ∠ 1 ① and x > &# 189; (x-3) ② is the root of the equation 2X-4 = ax about X, try to find the value of a?
Bonus points in ten minutes!
The title is wrong
2X + 38 ∠ 1 ① and x > ½ (x-3) ②
The results show that x ＜ - 18.5 and X ＞ - 3 have no real solution~
How can there be integer solutions?
Watchtower master, please correct the title
There are two points m and N on the straight line AB and AC on both sides of equilateral △ ABC. D is a point outside △ ABC, and ∠ MDN = 60 °, BDC = 120 ° BD = DC. Explore: when m and N move on the straight line AB and AC respectively, the quantitative relationship between BM, NC and Mn, and the relationship between the perimeter Q of △ amn and the perimeter l of equilateral △ ABC. (1) as shown in Figure 1, when m and N are on AB and AC, and DM = DN, the relationship between BM, NC and Mn is as follows What is the quantitative relationship between them___ In this case, QL = 0___ (2) as shown in Fig. 2, are the two conclusions of conjecture (1) still valid when DM ≠ DN on points m, n edges AB and AC? Write out your conjecture and prove it. (3) as shown in Figure 3, when m and N are on the extension line of edge AB and Ca respectively, if an = x, then q = 0___ (denoted by X, l)
(1) In this case, QL = 23. (2) conjecture: the conclusion is still valid. Prove: as shown in the figure, extend AC to e, make CE = BM, connect de. ∵ BD = CD, and ∵ BDC = 120 °, and ∵ DBC = ∵ DCB = 30 °. ABC is an equilateral triangle, and ∵ MBD = ≌ NCD = 90 °. In △ MBD and △ ECD: BM = CE ∵ MBD = ≌ ECD = DC ≌ MBD ≌ ECD (SAS) ）In △ MDN and △ EDN, DM = de ∠ MDN = ∠ EDN = DN, △ MDN ≌ △ EDN (SAS); △ Mn = ne = NC + BM. The perimeter of △ amn is q = am + an + Mn = am + an + (NC + BM) = (am + BM) + (an + NC) = AB + AC = 2Ab, while the perimeter of equilateral △ ABC is L = 3AB. ｜ QL = 2ab3ab = 23 If an = x, then q = 2x + 23L
If the solution set of the inequality ax ^ 2 - | x + 1 | + 2A ＜ 0 about X is an empty set, then the value range of a
When x ≤ - 1, the inequality ax & sup2; - |x + 1 | + 2A < 0 is ax & sup2; + X + 2A + 1 < 0
x> The inequality ax & sup2; - |x + 1 | + 2A < 0 is ax & sup2; - x + 2a-1 < 0
The solution set of ax & sup2; - |x + 1| + 2A < 0 is empty
A>0
1-4a(2a+1)
In △ ABC, ab = AC, BD bisects ∠ ABC intersects AC side at point D, ∠ BDC = 75 °, then the degree of ∠ A is______ .
Suppose the degree of ∠ A is x, then ∠ C = ∠ B = 180 − X2, ∵ BD bisection ∠ ABC intersection AC edge is at the point D ≠ DBC = 180 − x4, ∵ 180 − x2 + 180 − X4 + 75 = 180 °, ∵ x = 40 ° and the degree of ∠ a is 40 °. So the answer is: 40 °
The inequality x ^ 2 + ax + 1 ≥ 0 holds for all x ∈ (0,1 / 2), and the value range of a is obtained
The symmetry axis of F (x) = x ^ 2 + ax + 1 is x = - A / 2, f (0) = 1, and we can see its image by discussing the symmetry axis
In case 1, when - A / 2 = 0, f (x) is monotonically increasing in the interval of [0,1 / 2], f (x) > = f (0) = 1 > 0, so the condition a > = 0 is satisfied
The second case is 0-2
f(x)=X^2+AX+1
f(0)=1>0
There must be f (1 / 2) > = 0
If the axis of symmetry is in the middle, (a > - 1)
A ^ 2-4 is required