First, multiple-choice questions (3 points for each question, 9 questions, ***27 points)
1. The number of axially symmetric figures in the following figure is ()
1。
Axisymmetric figure of test point.
The analysis is based on the concept of axisymmetric graphics.
Solution: As can be seen from the figure, the first, second, third and fourth figures are all axisymmetric, and there are ***4 figures.
So choose D.
This topic examines axisymmetric graphics, the key of which is to find the axis of symmetry. After the graphics are folded along the axis of symmetry, the two parts can overlap.
2. The following operation is incorrect ()
A.x2？ x3=x5B。 (x2)3=x6C.x3+x3=2x6D。 (﹣2x)3=﹣8x3
Power of testing center and power of products; Merge similar projects; Multiplication with the same base.
The knowledge points examined in this question are the multiplication rule of the same base power, the multiplication rule of the power, the merger of similar items and the multiplication rule of products.
Solution: A, x2? X3=x5, correct;
B, (x2)3=x6, correct;
C, should be x3+x3=2x3, so this option is wrong;
D, (-2x) 3 =-8x3, correct.
So choose: C.
The knowledge points used in this topic are annotated as follows:
Same base powers's law of multiplication: the base is constant, and the exponents are added;
The law of power multiplication is: the base number is constant, and the exponent is multiplied;
Merging similar items only needs to add or subtract coefficients, and the letters and their indexes remain unchanged;
The power of the product is equal to each factor in the product multiplied by the power respectively.
3. The following judgment about the score is correct ()
A. when x=2, the value of is zero.
No matter what the value of x is, the value of x is always positive.
C no matter what the value of x is, it is impossible to get an integer value.
D. when x≠3, it makes sense.
The condition that the value of the test center score is zero; Definition of score; Conditions for meaningful scores.
The condition that the analysis score is meaningful is that the denominator is not equal to 0.
The condition that the fractional value is 0 is that the numerator is 0 and the denominator is not 0.
Solution: A. When x=2, the denominator x-2 = 0, and the score is meaningless, so A is wrong;
B, x2+ 1≥ 1 in the denominator, so the second formula must be established, so b is correct;
C, when x+ 1= 1 or-1, the value of is an integer, so c is wrong;
D, when x=0, the denominator x=0, and the score is meaningless, so d is wrong.
So choose B.
The condition that the value of the comment score is positive is that the numerator and denominator have the same sign, and the condition that the value is negative is that the numerator and denominator have different signs.
4. If the result of factorization of polynomial x2+mx+36 is (x-2) (x- 18), then the value of m is ().
A.﹣20B.﹣ 16C. 16D.20
Factorization of test sites-cross multiplication, etc.
Special calculation problems.
The result of factorization is calculated by multiplying polynomial by polynomial rule, and the value of m can be obtained by using the condition of polynomial equality.
Solution: x2+MX+36 = (x ﹣ 2) (x ﹣18) = x2 ﹣ 20x+36.
Available m =-20,
So choose a.
This topic reviews the factorization-cross multiplication, and mastering the method of cross multiplication is the key to solve this problem.
5. If the circumference of an isosceles triangle is 26cm and one side is 1 1cm, then the waist length is ().
A.11CMB.7.5cmc.11cm or above is incorrect.
Properties of isosceles triangle.
Split edge analysis 1 1cm is waist length and bottom edge.
Solution: ① When 1 1 cm is waist length, waist length is 1 1cm.
② When 1 1cm is the base, the waist length = (26- 1 1) = 7.5cm,
So the waist circumference is 1 1cm or 7.5cm
So choose C.
This question examines the nature of isosceles triangle, and the difficulty lies in discussing it in different situations.
6. As shown in the figure, in △ABC, AB=AC, ∠ BAC = 108, point D is on BC, BD=AB, and connected AD, then ∠CAD is equal to ().
A.30 B.36 C.38 D.45
Properties of isosceles triangle.
The analysis of ∠B and ∠BAD is based on the equality of the two bottom angles of the isosceles triangle, and then the solution can be obtained according to the calculation of ∠CAD =∠BAC∠BAD.
Solution: AB = AC, ∠ BAC = 108,
∴∠b=( 180 ﹣∠bac)=( 180 ﹣ 108)= 36，
BD = AB，
∴∠bad=( 180 ﹣∠b)=( 180 ﹣36)= 72，
∴∠cad=∠bac﹣∠bad= 108 ﹣72 = 36。
So choose B.
This topic examines the properties of isosceles triangle, mainly using the properties of isosceles triangle with equal base angles and equal corners. Memorizing nature and understanding maps accurately are the key to solving problems.
7. As shown in the figure below, △ Abe △ ACD, ∠ 1=∠2, ∠B=∠C are known, and the incorrect equation is ().
A.AB=ACB。 BAE =∠CADC. Bei =DCD. AD=DE
Test the properties of congruent triangles.
Analysis According to the nature of congruent triangles, congruent triangles's corresponding edges are equal, and congruent triangles's corresponding angles are equal, which can be judged.
Solution: ∫△ABE?△ACD, ∠ 1=∠2, ∠B=∠C,
∴AB=AC,∠BAE=∠CAD,BE=DC,AD=AE，
Therefore, A, B and C are correct;
The edge corresponding to AD is AE instead of DE, so D is wrong.
So choose D.
This paper mainly investigates the properties of congruent triangles, and it is the key to solve the problem to correctly determine the corresponding edge according to the known corresponding angle.
8. Calculation: (-2) 20 15? () 20 16 is equal to ()
A.﹣2B.2C.﹣D.
Test the power of the center and the power of the product.
In this analysis, the original formula is deformed directly by using the same base power multiplication algorithm, and the answer is obtained.
Solution: (-2) 20 15? ()20 16
=[(﹣2)20 15? ()20 15]×
=﹣.
So choose: C.
This topic mainly examines the multiplication operation of product and same base powers, and mastering the algorithm correctly is the key to solving the problem.
9. As shown in the figure, line A and line B intersect at point O, ∠ 1 = 50, point A is on line A, and there is point B on line B, so the triangle with points O, A and B as vertices is an isosceles triangle, and such point B has ().
1。
Determination of isosceles triangle in examination center.
According to the analysis that △OAB is an isosceles triangle, it is discussed in three situations: ① when OB=AB, ② when OA=AB, ③ when OA=OB, the corresponding point B is found and the solution is found.
Solution: To make △OAB an isosceles triangle, there are three situations that can be discussed:
(1) When OB=AB, make the middle vertical line of line segment OA, and the intersection of line segment OA and line segment B is b, so there is1;
(2) When OA=AB, take point A as the center, OA as the radius, and the intersection with line B. At this time, there is1;
(3) When OA=OB, make a circle with point O as the center, OA as the radius, and have two intersections with straight line B,
1+ 1+2=4,
Therefore, choose: d.
This topic mainly examines the nature of coordinates and graphics and the judgment of isosceles triangle; Classification discussion is the key to solve this problem.
Fill in the blanks (*** 10 small questions, 3 points for each small question, out of 30 points)
10. Calculate (|) | 2+(π | 3) 0 | | = 4.
Calculate the actual number of test sites; Zero exponential power; Negative integer exponential power.
Thematic calculation problems; Real number.
The first term of the original formula is calculated by the law of negative integer exponential power, the second term is calculated by the law of zero exponential power, the third term is simplified by the meaning of power, and the last term is simplified by the algebraic meaning of absolute value, and the result can be obtained by calculation.
Solution: The original formula = 16+ 1-8-5 = 4,
So the answer is: 4.
This topic reviews the operation of real numbers, and mastering the algorithm is the key to solve this topic.
1 1. If A-B = 14 and ab=6, then a2+b2=208.
Complete square formula of test center.
The analysis can be answered according to the complete square formula.
Solution: A2+B2 = (a-b) 2+2ab =142+2× 6 = 208,
So the answer is: 208.
This review examines the complete square formula, and the key to solving this problem is to memorize the complete square formula.
12. Given xm=6 and xn=3, the value of x2m﹣n is 12.
The division of the power of the same base in the examination center; Power and products.
This paper analyzes the division rule based on same base powers: the base number is constant, and the operation is enough by subtracting the exponent.
Solution: x2m ﹣ n = (XM) 2 ? xn = 36 ? 3 =12.
So the answer is: 12.
Comments This topic examines same base powers's knowledge of division and multiplication, which belongs to the basic topic. Mastering the algorithm of each part is the key.
13. When x= 1, the value of the score is zero.
The condition that the score of the inspection center is zero.
The analytical score of 0 is as follows: (1) molecule is 0; (2) The denominator is not 0. Both conditions must be met at the same time, and both are indispensable. Therefore, this problem can be solved.
The solution is x2 ~ 1 = 0, and the solution is x = 1.
When x =- 1, x+ 1=0, so it should be discarded.
So x= 1.
So the answer is: 1.
Commenting on this topic, it is investigated that the value of the score is zero, and two conditions need to be met at the same time: (1) the molecule is 0; (2) The denominator is not 0. These two conditions are indispensable.
14.( 1999? Kunming) It is known that the sum of the internal angles of a polygon is equal to 900, so the number of sides of this polygon is 7.
Test the inner and outer angles of the central polygon.
Answer the analysis according to the polygon internal angle and calculation formula.
Solution: Let the number of sides of a regular N-polygon be n,
Then (n-2)? 180 =900 ,
The solution is n=7.
So the answer is: 7.
This topic examines how to calculate the number of sides of a polygon according to its internal angle and calculation formula, and how to correctly calculate, deform and process data according to the formula when solving.
15. As shown in the figure, in ABC, AP=DP, DE=DF, DE⊥AB in E and DF⊥AC in F, the following conclusions are drawn:
①AD divides equally ∠ BAC; ②△BED?△FPD； ③DP∨AB； ④DF is the middle vertical line of PC.
The correct one is ① ③.
Congruent triangles's judgment and nature; The nature of the angular bisector; Properties of the vertical line in the line segment.
Thematic geometry problems.
According to the properties of the angular bisector, we get the AD bisector ∠BAC. Because the title doesn't give the condition to prove ∠C=∠DPF, we can't prove that △ bed △ FPD and DF are the perpendicular lines of PC according to the properties of isosceles triangle, so we can get ∠PAD=∠ADP first, and then ∠.
Solution: de = df, DE⊥AB in E, DF⊥AC in F.
∴AD divides∠∠ BAC equally, so ① is correct;
Because the title does not give the condition to prove that ∠C=∠DPF, only a right angle and an edge are equivalent, so it can't be proved that △ bed △ FPD and DF are the median lines of PC according to congruent triangles's judgment, so ② ④ is wrong;
AP = DP，
∴∠PAD=∠ADP，
∫AD splitting∠ ∠BAC,
∴∠BAD=∠CAD，
∴∠BAD=∠ADP，
∴DP∥AB, so ③ is correct.
So the answer is: ① ③.
Comments on congruent triangles's judgment and nature, the nature of bisector, the nature of perpendicular bisector, the nature of isosceles triangle and the judgment of parallel lines are comprehensive, but not difficult.
16. The number 0.00020 16 expressed by scientific notation is 2.0 16× 10 ~ 4.
Scientific notation of test sites-indicating smaller numbers.
Positive numbers whose absolute value is less than 1 can also be expressed by scientific notation, and the general form is a×10-n. Different from the scientific notation of large numbers, it uses negative exponential power, and the exponent is determined by the number of zeros before the first non-zero number from the left of the original number.
Solution: 0.0002016 = 2.016×10 ~ 4.
So the answer is: 2.0 16× 10 ~ 4.
Comment on this topic, scientific notation is used to represent smaller numbers, and the general form is a×10-n, where 1 ≤| a | < 10, and n is determined by the number of zeros before the first non-zero number from the left of the original number.
17. As shown in the figure, points A, F, C and D are on the same straight line, AF=DC, BC∨EF. A condition needs to be added to judge △ ABC △ def, and the added condition is EF=BC.
Congruent triangles's judgment of the test center.
The topic is open.
Analyze the added condition: EF=BC, then AC=FD can be obtained according to AF=DC, then ∠EFD=∠BCA can be obtained according to BC∠EF, and then △ ABC △ Def can be determined according to SAS.
Solution: additional conditions: EF=BC,
∫BC∨EF，
∴∠EFD=∠BCA，
∫AF = DC，
∴AF+FC=CD+FC，
That is AC=FD,
At △EFD and △BCA,
∴△EFD≌△BCA(SAS).
Therefore, EF=BC.
The comment on this topic mainly examines the judgment method of triangle congruence. The general methods to judge the coincidence of two triangles are SSS, SAS, ASA, AAS and HL.
Note: AAA and SSA cannot judge whether two triangles are congruent. When judging whether two triangles are congruent, there must be edges involved. If two angles are equal, the angle must be the included angle between two sides.
18. If x2 ~ 2ax+ 16 is completely flat, then a = 4.
The test center is completely flat.
Analyze the complete square formula: (AB) 2 = A2AB+B2, where the first two terms and the last two terms are the squares of the numbers X and 4, then the middle term is twice the addition and subtraction product of X and 4.
Solution: ∫x2∫2ax+ 16 is completely flat.
∴﹣2ax= 2×x×4
∴a= 4。
The comment on this topic is the application of the complete square formula. The sum of the squares of two numbers, plus or MINUS twice their product, constitutes a completely flat road. Pay attention to the sign of the double product so as not to miss the solution.
19. As shown in the figure, it is known that ∠ mON = 30, points A 1, A2, A3, … are ON ray ON, points B 1, B2, B3, … are on ray OM, △ A1B/kloc-.
Properties of equilateral triangle in test center.
General types of topics.
According to the properties of isosceles triangles and parallel lines, a 1b 1∨a2 B2∨a3 B3 and A2B2=2B 1A2, A3B3=4B 1A2=8, a4b4 = 8b1a2 =
Solution: ∫△a 1b 1 a2 is an equilateral triangle.
∴A 1B 1=A2B 1，
∫∠MON = 30，
OA2 = 4，
∴OA 1=A 1B 1=2，
∴A2B 1=2，
∵△A2B2A3 and △A3B3A4 are equilateral triangles.
∴a 1b 1∥a2b2∥a3b3,b 1a2∥b2a3，
∴A2B2=2B 1A2,B3A3=2B2A3，
∴A3B3=4B 1A2=8，
A4B4=8B 1A2= 16，
A5B5= 16B 1A2=32，
By analogy, the side length of △AnBnAn+ 1 is 2n- 1.
So the answer is 2n- 1.
This topic mainly investigates the properties of equilateral triangle and right triangle with an angle of 30. It is concluded from the conditions that OA5 = 2oa4 = 4oa3 = 8oa2 =16oa1is the key to solve the problem.
Third, answer the question (this big question is ***7 small questions, ***63 points)
20. Calculation
( 1)(3x﹣2)(2x+3)﹣(x﹣ 1)2
(2)(6x4﹣8x3)÷(﹣2x2)﹣(3x+2)( 1﹣x)
Mixed operations of algebraic expressions.
Analysis (1) is calculated by the rule of polynomial multiplication;
(2) Solving the problem by using algebraic mixed calculation rules.
Solution: (1) (3x-2) (2x+3)-(x-1) 2
=6x2+9x﹣4x﹣6﹣x2+2x﹣ 1
=5x2+7x﹣7；
(2)(6x4﹣8x3)÷(﹣2x2)﹣(3x+2)( 1﹣x)
=﹣3x2+4x﹣3x+3x2﹣2+2x
=3x﹣2.
This topic examines the mixed calculation of algebraic expressions. The key is to multiply each term of one polynomial by each term of another polynomial, and then add the products.
2 1. factorization
( 1)a4﹣ 16
(2)3ax2﹣6axy+3ay2.
Comprehensive application of common factor method and formula method in examination center.
Analysis (1) factorize the factor twice with the square difference formula;
(2) First extract the common factor 3a, and then continue to decompose the remaining polynomials by using the complete square formula.
Solution: (1) A4- 16
=(a2+4)(a2﹣4)
=(a2+4)(a+2)(a﹣2)；
(2)3ax2﹣6axy+3ay2
=3a(x2﹣2xy+y2)
=3a(x﹣y)2.
On this topic, we study factorization by putting forward common factors and formulas. If the polynomial has a common factor, first extract the common factor, and then decompose it by other methods, and at the same time decompose it thoroughly until it can't be decomposed.
22.( 1) Simplify the algebraic expression first, and then choose an A value that makes the original expression meaningful for evaluation.
(2) Solve the equation:
Simplified evaluation of test center score: solving score equation.
Thematic calculation problems; mark
This paper analyzes the general division of two items in brackets in the original formula (1), and uses the addition rule of fractions with the same denominator to calculate. At the same time, the simplest result is obtained by using the division law, and a=2 is substituted into the calculation to get the numerical value.
(2) Convert the denominator of the fractional equation into an integral equation, get the value of x by solving the integral equation, and get the solution of the fractional equation by testing.
Solution: (1) Original formula = [+]? =? =,
When a=2, the original formula = 2;
(2) After removing the denominator, 3x=2x+3x+3,
Transfer and merger: 2x =-3,
Solution: x =- 1.5,
X =- 1.5 is the solution of the fractional equation.
The key to solve this problem is to simplify the evaluation of the score and master the algorithm.
23. Establish a plane rectangular coordinate system as shown in the figure in a square grid composed of small squares with a side length of 1, and the grid triangle ABC is known (all three vertices of the triangle are on the small square).
(1) Draw the symmetrical triangle of △ABC about the straight line L△ A1b1:x =-1; And write the coordinates of A 1, B 1 and C 1.
(2) Find a point D on the straight line X =-L to minimize BD+CD, and the point D that meets the conditions is (-1,1).
Tip: The straight line X =-L is a straight line passing through the point (-1, 0) and perpendicular to the X axis.
Surveying and mapping of test site-axisymmetric transformation; Axisymmetric shortest path problem.
Analyze (1) to make the symmetrical points of points A, B and C about the straight line L: X =- 1, and then connect them in turn to write the coordinates of A 1, B 1 and C 1;
(2) Let point B 1 which is symmetric about X =- 1 connect CB 1, and the intersection with X =- 1 is point D. At this time, BD+CD is the minimum, and write the coordinates of point D. 。
Solution: The graph made by (1) is as follows:
A 1(3， 1)，B 1(0，0)，C 1( 1，3)；
(2) point B 1, where point b is symmetric about x =- 1,
Connect CB 1, and the intersection with X =- 1 is point d,
At this time, BD+CD is the smallest,
The coordinates of point D are (-1, 1).
So the answer is: (-1, 1).
Aiming at this problem, the drawing based on axisymmetric transformation is studied. The key to solve this problem is to make the positions of corresponding points according to the grid structure and connect them in turn.
24. As shown in the figure, AD is known to divide equally ∠CAE and AD∨BC.
(1) Prove that △ABC is an isosceles triangle.
(2) When ∠CAE is equal to how many degrees, is △ABC an equilateral triangle? Prove your conclusion.
Determination of isosceles triangle of test center; Determination of equilateral triangle.
Analysis (1) According to the definition of angular bisector, we can get ∠EAD=∠CAD, and then ∠EAD=∠B, ∠CAD=∠C according to the properties of parallel lines, and then we can get ∠ B = ∠.
(2) According to the definition of angular bisector, we can get ∠ EAD = ∠ CAD = 60, and then according to the properties of parallel lines, we can get ∠ EAD = ∠ B = 60, and then we can get ∠ B =.
Solution (1) Proof: ∫AD Dichotomy ∠CAE,
∴∠EAD=∠CAD，
∫ AD ∨ BC,
∴∠EAD=∠B,∠CAD=∠C，
∴∠B=∠C，
∴AB=AC.
So △ABC is an isosceles triangle.
(2) Solution: When ∠ CAE = 120, △ABC is an equilateral triangle.
∫∠CAE = 120, AD bisection ∠CAE,
∴∠EAD=∠CAD=60，
∫ AD ∨ BC,
∴∠EAD=∠B=60，∠CAD=∠C=60，
∴∠B=∠C=60，
△ ABC is an equilateral triangle.
The judgment of isosceles triangle, the definition of angle bisector, the properties of parallel lines and the simple memory of properties are the key to solving problems
A factory now produces an average of 50 more machines a day than originally planned. It takes the same time to produce 600 machines now as planned to produce 450 machines. How many machines are produced on average every day?
Application of score equation of test center.
Special application problems.
This paper analyzes the ability of fractional equation to solve practical problems. Because the time to produce 600 machines now is the same as the time to produce 450 machines originally planned, the equivalence relationship can be obtained as follows: the time to produce 600 machines now = the time to produce 450 machines originally planned.
Solution: Assuming that X machines are produced every day on average, it is originally planned to produce (x-50) machines.
According to the meaning of the question:
Solution: x=200.
Test: when x=200, x(x﹣50)≠0.
∴x=200 is the solution of the original fractional equation.
A: At present, an average of 200 machines are produced every day.
Comments: It is the same to solve application problems with column fractional equation and with full column equation. The key point is to find out the equation relationship accurately, which is the basis of the column equation. The difficulty lies in the analysis of the known conditions of the problem, that is, the examination of the problem. Generally speaking, there are two conditions for application problems, one is explicit, which is given directly in the problem, and the other is implicit, which is given by the implicit condition of the problem. In this question, "now, on average, 50 more units are produced every day than originally planned."
26. As shown in the figure, △ACB and △ADE are isosceles right triangles, ∠ BAC = ∠ DAE = 90, and points C, D and E are on the same straight line, connecting BD. Verification:
( 1)BD = CE；
(2)BD⊥CE.
Congruent triangles's judgment and nature; Isosceles right triangle.
Special topic proof questions.
Analysis (1) Through conditional proof △ bad △ CAE, we can draw a conclusion;
(2) According to the properties of congruent triangles, we can get ∠ Abd = ∠ Ace+∠ DFC = 90 and ∠ FDC = 90.
This solution proves that (1)∫△ACB and △ADE are isosceles right triangles.
∴AE=AD,AB=AC,∠BAC=∠DAE=90，
∴∠BAC+∠CAD=∠EAD+∠CAD，
That is ∠BAD=∠CAE,
In △BAD and △CAE,
,
∴△BAD≌△CAE(SAS)，
∴bd=ce；
(2) As shown in the figure,
∫△BAD?△CAE，
∴∠ABD=∠ACE，
∫∠CAB = 90，
∴∠ABD+∠AFB=90，
∴∠ACE+∠AFB=90，
∠∠DFC =∠AFB，
∴∠ACE+∠DFC=90，
∴∠FDC=90，
∴BD⊥CE.
Comments This topic examines congruent triangles's judgment and property application, vertical judgment and property application, the property application of isosceles right triangle, the application of Pythagorean theorem, and the key to solving problems is to use congruent triangles's property.
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