1-4

5、2c； 6、 16cm； 7 、( 1)AB； (2) CD; (3)EF； (4)3cm； 3cm8、 10

9 questions (1)20 degrees; (2)2/2 1∠EAD=∠C-∠B The reasons are as follows: (1) shows that ∠ EAD = ∠ EAC-∠ DAC =1/.

Substitute ∠BAC = 180-∠ B-∠ C into ① to get EAD = half ∠ C-half ∠ B, so 2 ∠ EAD = ∠ c-∠ b.

1 1.2. 1 interior angle of triangle

1、90 ; 2、50 ; 3、 100 ; 4、20 ; 5、300 ; 6、50 ; 7、B 8、C； 9 questions 120 degrees; 10 question 36 degrees; 1 1 problem (1)75 degrees; ②75 degrees.

1 1.2.2 outer corner of triangle

1, B 2, C 3, C 4, C 5, 180 degrees; 6. 1 15 degrees; 7 solution: Because ∠ADC=∠B+∠BAD. ∠DAC =∠EAC+∠DAE； Because ∠ADC =∠DAC；; ∠B=∠EAC, so∠ ∠BAD=∠DAE, so AD shares∠ ∠BAE equally.

Question 8 (1)2∠F=∠B+∠D

(2)x=3

1 1.3 sum of polygons and their internal angles

1, A 2B 3C 4B 5A 6D 7 Question 8; 8 questions 120 degrees; 40 degrees; 80 degrees; 120 degrees 9 questions 900 degrees; 10 question 9; 1 1 540 degrees; 12 question 540 degrees; The problem 13 proves that in quadrilateral ABCD, ∠A=∠C=90 degrees, so ∠ABC+∠CDA= 180 degrees, because ∠ 1 = ∠ 2, Because ∠A=90 degrees, ∠ 1+∠AEB=90 degrees. Because ∠ 1=∠2, ∨AEB =∠3. So be Σ FD

Chapter XII

12. 1 full triangle

1, A 2, B 3, B 4, D 5, 120 degrees 6 (1) prove that triangles are similar; (2) Proof of similarity and chamfering calculation; (7) Prove the similarity by chamfering the given angle, the angle DAE is 73 degrees and the angle CEF is 34 degrees.

12.2 Determination of Triangular Congruence

The first category: 1, D 2, D 3, 2 4, AE=AD 5, proving congruence 6, proving congruence 7 with conditions, proving triangle congruence and chamfering.

The second category: 1, C 2, AB=AC 3, AE=AC 4, AE=AF Prove congruence 5 with conditions, prove congruence (SAS)6, (1), prove two triangle congruences (SAS) (2) with E as the midpoint, and rotate (3) to prove triangle congruence.

The third category: 1, D 2, D 3, not AC=DF or angle B= angle e or angle A= angle d (the answer is not unique) 4. Ditto is not unique. 5. Use parallel relation to prove congruence. 6.( 1)AB=AD (the answer is not unique) (2) slightly 7. Congruence (AAA)

The fourth category: 1, C 2, B 3, just one. 4.( 1) Short (HL) (2) The known angle ACF is 60 degrees.

5. Use ABCD as a square to prove that the triangle is congruent and the sides are equal. 6.DE=DF (or CE//BF or angle ECD= angle DBF or angle DEC= angle DFB, etc. ) omitted.

Properties of bisector intersecting 12.3

The first category: 1, D 2, D 3, 6 4, vertical 5, 8 6, with congruent inverted edge AB=4cm 7, omitted.

The second category: 1, D 2, B 3, 90 degrees 4, slightly 5. Only two triangular congruences are used to prove 6, and triangular congruences (AAS) are used and then substituted into the same amount.

Chapter 13 13. 1. 1 axisymmetric

1, D2D3①②③4, axisymmetric, 5; 5 questions 4 in turn 4; 6 questions 10cm square; Question 7 ∠ABC=60 degrees, ∠C=30 degrees.

Properties of the perpendicular line of 13. 1.2 line segment

1 and D2B3B4 are on the vertical line of BC.

5 solution: AB+BD=DE

Proof: Because AD ⊥ BC and BD = DC, point A is on the middle vertical line of BC. So AB=AC. Because point A is on the middle vertical line of AE, AC=EC. So AB=EC. Because BD = DC, DE = CD+CE, DE=AB+BD.

6 solution: (1) AC = BC = AD = BD, OC = OD, OA = OB.

(2)OE=OF。 Because AB bisects CD vertically, angle AOC= angle AOD and OC=OD. In AOC triangle and AOD triangle,

AO=AO，

∠AOC=∠AOD

OC=OD

So triangle AOC≌ triangle AOD(SAS). So ∠ Cao = ∠ Tao, and, so oe =.

7 Question Angle B=30 degrees

13.2 the first lesson of drawing axisymmetric graphics

1、D； 2、D； 3. Omission; Question 4 (1) is omitted; (2) congruent triangles has: △ ACB △ AEB, △ ACD △ AED, △ BCD △ Bed.

Five themes

The second category is 1A, 2(2, -3). 3(-2, -3), vertical; 4 topics are abbreviated. A 1(0，2)，B 1(-3，-5)C 1(5，0)

5 questions (1) sketch. (2) If the coordinate of P 1(x 1, y 1) is p2(x2, y2), then x 1+x2=2, y 1=y2.

Question 6 (1)a= negative 4/5 and b= negative 3/5. (2) A+B = negative 7/5.

13.3 isosceles triangle 13.3. 1 isosceles triangle of the first kind

1、D2、D； 3B； 4C； 5 questions 36 degrees

6 solution: △ Abe △ ACD, △ BCD △ CBE or △ BFD △ CFE (just write two). (Select a group of triangles for brief proof of congruence)

Question 7 (1) proves that in triangle ACD and triangle ABE,

∠A=∠A，∠ADC=∠AEB=90，AC=AB

So triangle ACD≌ triangle ABE

So AD=AE.

(2) Solution: perpendicular to each other. Reason: In RT△ADO and RT△AEO, because OA = OA and AD = AE, RT△ADO≌TR△AEO∠ Dao =∠EAO, that is, OA is the bisector of ∠BAC. And because AB=AC, so OA⊥BC.

Second lesson

1、B； 2、A； 3、 10cm； 4、3; 5.△ABC (or △BDC or △DAB) (abbreviation proof)

6 solution (1)AD=AE. The reasons are as follows: ∠E=∠C=90, ∠ BDF+∠ B = 90 because ∠ BDF+∠ Ade.

(2) Omission

Question 7 (1) proves that ∠ OB=OC ∠ OCB. Because BD and CE are higher than △ABC, ∠ BDC = ∠ CEB = 90.

In RT△BDC, ∠ BCD = 90-∠ DBC;

At RT△CEB, ∠ BCD = 90-∠ ECB. So ∠BCD=∠CBE So AB=AC, that is, △ABC is an isosceles triangle.

(2) Solution: O point is on the bisector of BAC. The reasons are as follows: in RT△ABD and TR△ACE, ∠ A = ∠ A, ∠ ADB = ∠ AEC = 90, AB=AC, so TR△ABD ≌ RT△ACE. So BD=CE. Because OB=OC, OD =

13.3.2 equilateral triangle

1、D； 2、 17; Question 3 proves that CD=2AD in RT△ADC because AB = AC, ∠ BAC = 120, ∠ B = ∠ C = 30. Because ∠ BAC = 120, ∞.

Question 4 proves that since triangle ABC and triangle DEC are equilateral triangles, ∠BCD=∠ACE, CD = CE, ∠ ABC = ∠ BCA = ∠ ECD = 60, ∠ BCA-∠ DCA = ∠ ECD.

Answer to Question 5 (1) Sketch (2) proves that BD bisects angle ABC because triangle ABC is an equilateral triangle and d is the midpoint of AC. So the angle ABC = 2- the angle DB=DE. Because CE=CD and angle CED= angle CDE. Because angle ACB= angle CED+ angle CDE, angle ACB = 2- angle CED.

BM=EM (three in one)

Chapter 14 Multiplication of 14. 1. 1 same base powers

1、A2、C3、C4、D

5 questions 5; 6 questions 40; 7 questions 2 to the 99th power; 8 questions (1) 0; (2)0.9 question x = 5；; 10 problem m=3.

14. 1.2 power

1、D； 2、B； 3、C； 4、C； 5、0; 6、2; 7, 2 times x 12 power; 8 questions 2187; 9 questions 30625; The sixth power of 10 problem (1)2; 2 to the 6th power; 3 to the 6th power; 3 to the 6th power; The third power of the second power of bracket 3 = the second power equal to the third power of bracket 3; 4 to the 6th power; 4 to the 6th power; Cubic minus 4 of brackets = quadratic minus 4 of brackets = cubic of brackets.

(2) the n power of the m bracket of A = the m power of the n bracket of A. ..

③ the cubic of a.

The power of the product of 14. 1.3

1、D； 2、C； 3 questions m=3, n = 2；; 4 Question 5 Square 5 Cubic 5 quintic 5 quintic square brackets less than 5 quintic square brackets less than 5 quintic square brackets less than 5 quintic square brackets less than 5 quintic square brackets; Question 5: The 26th power of A times the 30th power of B; 6 questions 243; 7 questions 2 16.

14. 1.4 Algebraic expression multiplication of the first kind 1, c; 2、D； 3、A； 4、D； 5、C； 6 questions ② ⑤; Question 7-12xde6 times Y7; 8 questions 3/8 π times the quadratic power of y; Question 9, 2.

The second category is 1, a; 2、D； 3、B； 4 square of problem 4a +7ab- 15b; 5 questions-6; 6 questions12; Question 7 (1) cubic of m+cubic of n; (2) 2) The cubic of 2y+the quadratic of 3y -2 1y+5.

Question 8 simplifies the original formula to 6(2n- 1). No matter what the nonzero natural number n is, 2n- 1 is an odd number, so the values of algebraic formulas n (n+7)-(n-3)-(n-2) can be evenly divided by 6. Question 9 proves that the original formula = 12, so it has nothing to do with the value of X.

10 question 77;

Question 1 1 (1) Calculate the square of the cross-sectional area of 1a+ 1ab, and then draw a conclusion.

(2)(50a squared +50ab).

The third category is 1, b; 2、C； 4、C； Question 5 -4a Party B; Problem 6 -(m-n) cubic; 7 questions12; 8 questions (1)2ab quadratic; (2) 2) that b3 pow of the 7a party+the 4th power of the 6a party; You can buy 100 pens and 300 notebooks for 9 questions.

14.2 multiplication formula 14.2. 1 square difference formula

1、C； 2、A； 3、A； 4、C； 5、B； 6、A； 7 questions (1) 9964; (2)5x square -5y square'; (3) the fourth power of X-the fourth power of Y. Question 8-6; 9 Solution: Yes, for the following reasons: A= 15n squared. Because n is a positive integer, the square of 15n must be divisible by 15. 10 problem 2 to the power of 128.

14.2.2 complete square formula, class 1, class 1, d; 2、D； 3 question 2; 4 questions 4b; 4a Party; 16ab .514; 6 questions-2; 7 questions (1) 45; (2)57; 8 questions124; Nine questions, 8xy.

Category II: 1, b; 2、C； 3 questions 6x; 4 questions 4y square -9x square -z square-6XZ; Question 5: A+B+C+2AB-2BC-2AC; 6 questions 4yz；; 7 questions are greater than; Greater than; Greater than; =; The rule is that Party A+Party B is greater than or equal to 2ab. It is proved that A+B is greater than or equal to 2ab because the square of A -2ab+b =(a-b) is greater than or equal to 0.

Questions 8 and 7.

14.3 factorization 14.3. 1 common factor method 1, c; 2、C； 3、B； 4、D； 5 questions (X-Y) (A+B+C); 6 questions x squared (2x-1); 7 question 3 (m+n) (m-n); 8 questions 9a+3a-2; Questions 9-6; 10 problem 14 1.3m square.

Formula method 14.3.2 class 1, 1, d; 2、B； 3、D； 4 questions-198000; Question 5200.5-12; 6 questions 3m (2x-y+n) (2x-y-n); 7 questions (1) x (x+3) (x-3); (2)m squared (m+n) (m-n); (3)-3(p-q)(p+q)； 8 Question A (A+3) (A-3); Question 9 (x+4y) (x-4y); 10 question 5050.

Category II: 1, c; 2、A； 3、D； 4 question b (a+1); Question 5 (1) 10 to the sixth power; (2)4;

6 questions x = y.

Chapter 15 Score 15. 1 score 15. 1 from score to score 1, B2, D3, D4; 5 questions (1)8x, 80% of x; (2)a+b； a-b； (3) 4 x-Y. Question 6 (1) According to the meaning of the question, it is concluded that 2x-4 is not equal to 0 and x is not equal to 2. Therefore, when x is not equal to 2, the original formula is meaningful. (2) According to the meaning of the question, x-3=0 and the solution x=3. So when x=3, the original formula value is 0. (3) According to the meaning of the question, x-3=2x-4. Solution x= 1 So when x= 1, the original fraction value is positive. (4) According to the meaning of the question, it is found that x-3 is greater than 0 and 2x-4 is greater than 0. Found that x is greater than 3. Or x-3 is less than 0, 2x-4 is less than 0, and x is found to be less than 2. Therefore, when x is greater than 3 or less than 2, the value of the formula is positive. (5) According to the meaning of the question, x-3 is greater than 0, and 2x-4 is less than 0. There is no solution. Or x-3 is less than 0 and 2x-4 is greater than 0. It is found that 2 is less than x and less than 3. So when 2 is less than x and less than 3, the value of the score is negative. Question 7 (1)-A's 10 power points to B's 29th power. (2) (-1) of n+1times the n power a of 3n- 1 times.

Basic properties of the score 15. 1.2 class 1, B2, C3, D4, C and question 5 (1)1; (2) One third of x +3.6 The original formula = two thirds of x+y (x-y); -2.7 out of 5.

Category II: 1, c; 2、C； 3、A； The simplest common denominator of question 4 (1) is the concrete division of 6x square y .. (2) The simplest common denominator is (x+2)(x-2). The simplest common denominator of question 5 (1) is 12a, the abbreviation of bc; (2) The simplest common denominator is 2(y+2)(y-2). 15.2 fractional operation 15.2. 1 fractional multiplication and division 1, 1 A2, C3, A4, an of bm in class. 5 questions (1)yc. (2)-x and y. (3) 1. (4) x+3y of X+Y. Question 6-1.7 (1)-y divided by the 5th power of 2, X divided by the 3rd power of Y = the 2nd power of Y, and the 7th power of the 3rd power of X divided by (5th power of Y -x power) ... If any score is divided by the previous time, the result is the 2nd power of -y. Question 5 (1)a-b to the fourth power of B (2)a's b-a6 question A-65438+2 full marks; Question 7 a- 1 a+ 1.

15.2.2 addition and subtraction of scores 1 question 0; 2 questions x- 1 min 1.3 questions c; 4 Original formula = A+2/2; 2/5

The second kind 1, two points of A2D31; 4 questions x+65438+4/0; Question 5 m-6.6 Original formula = 2A+8; 6; 7 Original formula = x-2x+1; X satisfies that -2 is less than or equal to x is less than or equal to 2 and is an integer. If the score is meaningful, x can only take 0, -2. When x=0, the original formula =-1; When x=-2, the original formula = 1.8. After simplification, the original formula is x+ 1, less than -4 and less than -2, so x=-3. When x=-3, the original formula =2.

15.2.3 Integer exponential power 1 class hour 1, C2, A3, D4-8.5 A square problem (-3) 1-n power X n power; 6 questions 0; 7 questions 5; Eight questions. eight. The second category, 1 question 2.3 times 10 to the -6 power; 2 Question 2. 1 times 10 to the fifth power; 3 question c; 4 the simplified original formula = 1 of a and-2 of 3; 15.3 fractional equation class 1 problem1x =-3; 2 problem m is greater than -6, and m is not equal to-4; 3 question d; Question 4: X = 0 of 65438+3. It is verified that X = 1 of 3 is not the solution of the original equation, and the original equation has no solution. Question 5: X = 5/2, which proves to be the solution of the original equation. Question 6 x=6, which is the solution of the original equation after the test. If the root of the problem has the meaning of the problem, you can get 1-x=3 in 2-x, and the solution is x = 5 in 2. It is verified that x = 5/2 is the solution of the original equation. The second category, 1 question a; 2 questions c; Question 3 Workshop A can process 384 pieces per day, and Workshop B can process 320 pieces per day. Question 4 (1) Every pencil is purchased in 4 yuan for the first time. (2) The price of each set is at least 6 yuan. The average speed of Xiaoli's return is 75 kilometers per hour. Question 6: The garrison in this area was originally reinforced by 300 meters every day. 7 the date specified for the topic is x days. According to the meaning of the topic, it is necessary

X = 3+(x+6)= 1。 X=6 is the solution of the listed fractional equation. Obviously, scheme (2) is irrelevant. Scheme (1) 1.2*6=7.2 (ten thousand yuan) Scheme (3) 1.2*3+0.5*6=6.6 (ten thousand yuan), because 7.2 > 6.6, we choose it without delaying the construction period.