1.a； 2.b； 3.a； 4.c； 5.d； 6.B
Fill in the blanks (4 points for each small question in this big question, out of 48 points)
7.； 8.； 9.； 10.； 1 1..； 12.
13.； 14.； 15.36; 16.； 17.2 or 8; 18. 1
Iii. Answer questions (19~22 is 10, 23~24 is 12, and 25 is 14, out of 78).
19. solution: the original formula =-3 points.
= - .
= - .
When, the original formula =-.
= - .
20. Solution: Multiply both sides of the equation to get:-2 points.
Namely:-3 points.
So,-4 points.
After inspection, I gave up in order to increase my roots. -1 min.
So the solution of the original equation is
2 1. solution: a is AD⊥BC in D-.
Tanb =, ∴ If AD= k, BD=4k,-2 points.
∵∠C=600，∴DC=，AC = -。
∫ BC = 12, ∴∴-.
∴ AC = -。
22.( 1) 100 - .
(2) four-.
(3) 36 - .
(4) 1900 - .
(5) no. Because the height of some seventh-grade students does not represent the height of first-grade students. -2 points
23. It is proved that (1)∵M is the midpoint of AB, and AD⊥BC, BE⊥AC,
∴2 points, 2 points.
∴ me = MD，-。
△ med is an isosceles triangle-66.
(2) ∵∴∠ Mei = ∠ MEA,-1 min.
∴∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠873
Similarly, ∴∠ MAD = ∠ MDA,-1 min.
∴∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠873
EMD =∣EMD
= 2 ∠ Mae-2 ∠ MAD = 2 ∠ DAC-65438。
24. solution: (1)∫ straight line passes through ∴∴ k = point c-1 point.
∴A(-2,0)，-。
The parabola intersects with point A and point C of∴.
That is, the analytical formula of ∴ parabola is-2 points.
(2) Vertex D-can be found.
Let DH⊥y axis intersect with Y axis at h-point.
∴ - 6544
= - .
25. (The full mark of this question is 14, item (1) is 4, item (2) is 6, and item (3) is 4).
Solution: (1) △ ABF ∽△ GBC, △ FDE ∽△ CGE ∽ △ BCE-4 points.
(2) Method 1:∵ divided into equal parts ∠B, ∴∠ABE=∠EBC,
∵ad//bc,∴∠afb=∠ebc,∴∠abe=∠afb,∴ab=af，
∴AF=4，DF= 1，-。
∵ AD//BC, ∴DF:BC=DE:EC, ∴ DE =, CE =-2 points.
∵AD//BC,AB=CD,∴∠BCD=∠ABC
* CG shares ∠BCD, BE shares ∠ABC, ∴ CBG = ∠ BCG, ∴BG=CG.
Let BG = CG = x, then it is given by △FDE∽△CGE, DF:CG=DE:GE, ∴ Ge =-1.
And from △CGE∽△BCE, EC2=EG? EB is
∴, which means BG =-2 points.
Method 2: Find DF= 1,-65438+.
Get DE =, CE =-2 points.
Let DF:BC= 1:5 let EF=x, BE=5x, from △FDE∽△CGE, score-1.
And from △CGE∽△BCE, EC2=EG? EB, that is, get- 1
Get-1 minute.
(3)① When the time is right, point A is within ⊙ P ..-2 points.
(2) When point A is within ⊙P and point E is outside ⊙P-2.