1. Decomposition factor: 8a2-2 =.
2. simplification; When m =- 1, the value of the original formula is.
3.2065438+In February 2002, the State Council issued the newly revised Ambient Air Quality Standard, which increased the monitoring of PM2.5 for the first time. PM2.5 refers to particulate matter with a diameter less than or equal to 0.000 0025m in the atmosphere, and 0.000 0025 is expressed by scientific counting method as follows
.
4. If the sum of the internal angles of a polygon is 1080, then the number of sides of the polygon is.
5. In order to make the quadratic radical meaningful, the range of x is.
6. When you roll a hexahedral dice, the probability of 1 point is, the probability of 7 points is, and the probability of not exceeding 6 points is.
7. It is known that points E, F, G and H are the midpoints of sides AB, BC, CD and DA of quadrilateral ABCD, respectively. If AC⊥BD and AC≠BD, the shape of the quadrilateral EFGH is (fill in "trapezoid", "rectangle" or "diamond").
8. Calculation:++=,
2. Multiple choice questions (3 points for each question, * * * 24 points)
9. The following factorization error is ()
a . 3a(a-b)-5b(a-b)=(a-b)(3a-5b)b . 4x 2-y2 =(2x-y)(2x+y)
c-4x 2+ 12xy-9 y2 =-(2x-3y)2d . x4-8x2y 2+ 16y 4 =(x2-4 y2)2
10. The following calculation error is ()
University of California, USA.
The condition that a 1 1. quadrilateral cannot be judged as a parallelogram is ().
A. two groups of opposite sides are parallel respectively. B. One set of opposite sides is parallel and the other set is equal.
C. a group of opposite sides are parallel and equal. The two groups of opposite sides are equal respectively.
12. The following simplification is correct ()
A. when x ≥1=1-x B. when a ≥ 0 and b ≥ 0 = 14 ab2.
C.=5+2 D。
13. The winning probability of a lottery ticket is 1%. The following statement is correct ().
A. If you buy a lottery ticket like 1, you won't win the prize.
B. If you buy a lottery ticket like 1, you will definitely win the prize.
C. If you buy 100 such lottery tickets, you will definitely win the prize.
D. When a large number of lottery tickets are purchased, the winning frequency is stable at 1%.
14. The result of solving the fractional equation is ()
A. No solution B.C.D. 1
15. As shown in the figure, in the isosceles trapezoid ABCD, AD∨BC and diagonal AC. BD intersects at point o,
The following conclusion is not necessarily correct ()
A.AC=BD B.OB=OC
C.∠BCD =∠BDC d∠ABD =∠ACD
16. As shown in the figure, the circumference of the diamond-shaped ABCD is 24cm, and the diagonal AC and BD intersect.
At point o, e is the midpoint of AD, connecting OE, and the length of line segment OE is equal to.
A.2 cm B 2.5 cm C 3 cm. 4 cm in diameter
3. Operation questions (5 points for each small question, 20 points for * * *)
17. Factorization:
18. Solve the fractional equation:.
19. Simplify first, and then choose an appropriate integer from the range of as the value of.
20. Given x=, y=, find the value of x2-xy+y2.
4. Application problem (8 points for this big problem)
2 1. Li Ming went to the school 2. 1 km away from home to attend the third grade party. When he got to school, he found that the performance props were still at home. At this time, there are still 42 minutes before the party starts, so he immediately walks home at a constant speed, takes 1 minute to get the props at home, and then immediately rides his bike back to school at a constant speed. As we all know, it takes Li Ming 20 minutes to go to school by bike than to walk home from school, and
(1) What is Li Ming's walking speed (unit: m/min)?
(2) Did Li Canming get to school before the party started?
Verb (abbreviation of verb) proves and asks questions (8 points for each small question, * * * 24 points)
22. As shown in the figure, diagonal AC and BD of quadrilateral ABCD intersect at point O, BE⊥AC intersects at point E, and DF⊥AC intersects at.
Point o is the midpoint between AC and EF.
(1) verification: △ BOE △ degree of freedom;
(2) If OA= BD, what special quadrilateral is quadrilateral ABCD? Please explain the reason.
23. As shown in the figure, it is known that the diagonal of rhombic ABCD intersects at point O, extending AB to point E, making BE=AB and connecting CE.
(1) verification: BD = EC
(2) If ∠ e = 50, find the size of ∠BAO.
24. It is known that the square ABCD, AC and BD intersect at point O, the right-angled vertex of the triangle coincides with O, and its two right-angled sides intersect with AB and BC at points E and F respectively.
(1) When the triangle rotates around point O until OE is perpendicular to AB (as shown in figure 1), it is verified that BE+BF= OB.
(2) When the triangle rotates a (0 < a < 45) counterclockwise around the O point under the condition of (1), as shown in Figure 2, does the above conclusion hold? If yes, please give proof; If not, please explain why.
Shuangfeng County 20 13 Last semester Grade 8 Mathematics Volume II Final Exam Reference Answers and Grading Criteria.
1. Fill in the blanks (3 points for each small question, ***24 points)
1.2(2a+ 1)(2a- 1)2. 1 3.2 . 5× 10-6 4.8 5 . x≤6.0， 1 7。
2. Multiple choice questions (3 points for each question, * * * 24 points)
9.D 10。 C 1 1。 B 12。 D 13。 D 14。 A 15。 C 16。 C
3. Operation questions (5 points for each small question, 20 points for * * *)
17. solution: the original formula 1 min.
3 points
5 points
18. Solution: After removing the denominator, 3x+x+2 = 4,2 points.
Solution: x=, 3 points
X= is the solution of the original equation. Five points.
19. Solution formula =
Because, let x=0 and the original formula =- 1.
20 solution: because x =- 1, 1 min.
Y = = 7+4 = 2 points
5 points
4. Application problem (8 points for this big problem)
2 1. Solution: (1) If the walking speed is m/min, the speed of the bicycle is m/min. 1 min.
According to the meaning of the question: 3 points
Get 4 points
Proved to be the solution of the original equation, 5 points.
Li Ming's walking speed is 70 meters/minute. Six minutes.
(2) According to the meaning of the question: 7 points.
Li arrived before the party started. 8 minutes [
Verb (abbreviation of verb) proves and asks questions (8 points for each small question, * * * 24 points)
22. solution: (1) proof: ∫be⊥AC. df⊥AC, ∴∠ beo = ∠ dfo = 90. 1 point
Point o is the midpoint of of ef 2 minutes.
∠∠DOF =∠∴△boe≌△dof(asa). BOE scored 3 points.
(2) The quadrilateral ABCD is a rectangle. 4 points
The reasons are as follows: ∫△ BOE△ degree of freedom, ∴OB=OD.5 points.
Oa = oc, ∴ quadrilateral ABCD is a parallelogram. 6 points
oa = bd，OA= AC,∴BD=AC。 7 points
The parallelogram ABCD is a rectangle. 8 points
23. solution: (1) proof: ∵ diamond ABCD,
∴AB=CD, ab∨CD, 1 min.
And ∵BE=AB,
∴BE=CD, be ∑CD, 2 points.
∴ Quadrilateral BECD is a parallelogram with 3 points.
∴bd=ec； 4 points
(2) solution: ∵ parallelogram BECD,
∴BD∥CE, 5 points.
∴∠ ABO =∠ E = 50, 6 points.
And: diamond ABCD,
∴ AC? BD, 7 points
∴∠ Bao = 90 ∠∠ ABO = 40.8 points.
24.( 1)∵ABCD is a square, and O is the intersection of diagonal AC and BD.
ob⊥oc,bc=·∴ob=oc。 1.
And ∵OE⊥AB, OF⊥BC,
∴OE=OF
∴Rt△BOE≌Rt△COF 2 points
∴BE=CF 3 points
∴BE+BF=CF+BF= OB。 Four points.
(2)BE+BF= OB still holds. Five points.
The reasons are: ∫≈EO b+∠BOF = 90, ∠ COF+∠ BOF = 90.
∴∠EOB=∠COF 6 points
ob = oc，∠ OBE = ∠ OCF = 45。
∴△BOE≌△COF 7 points
∴BE=CF
∴ Be+BF = CF+BF = ob.8