Winter vacation homework, the eighth grade math, Volume One answers the new classroom math! New classroom holiday life! Beijing publishing house and Beijing normal university edition cover mountains a
Winter vacation homework, the eighth grade math, Volume One answers the new classroom math! New classroom holiday life! Beijing publishing house and Beijing normal university edition cover mountains and rivers and enterprises
These questions start from the first page and go down. The order is arranged, and there is no guarantee of right or wrong ... ~ Math 1, DACB 2, AC=DF 45cm 3, 1. ∫ao shares ∠BAC∴do = EO∴∠DOB =∠EOC∫CD⊥ Abe ⊥ AC ∴ ODB = ∠ OEC = 97. ∠ A =∠ ACB = 60 ∴△ ACD △ BCE (SAS) ∴△ ACD = ∠ CBE ∠△ ABC is an equilateral triangle ∴△ACD+∞. ∠ BFC =180-∠ CBE-∠ BCD =120 I, DDDA II, BD = B'd' 180 III, (1)∑△ABC is equal. 1.DDC II, 30 3cm 90 III, (1) Using HL theorem, we can get △ ace △ CBD ∴ AE = CD (2) BD = CE =1/2bc = 6cmi, CDCB II, ∠ ADC. The vertical foot is g, h is ∠ gad+∠ adg = 90, ∠ had+∠ dah = 90, that is ∠ BAC+∠ gdh =180 ∠ BAC+∠ EDF. e = DF ^ 2。 According to symmetry principle I, DCCC II, the circumference of congruent symmetry axis 3cmb (-2,2) c (-2,2 2) d (2 2,2) 470 or 55 III, ∵△ABE is 10, that is, △AEC+Be+AB = 650. ED⊥AC ∴ED is the vertical line of the triangle AEC, that is, △AEC is an isosceles triangle ∴ae=ec∶ca = CB ∴ac+ab= 10 and ∶AC-ab = 2∴AC = 6 ab = 4. 0 15 three, 1. ∫de bisects AB ∴AD=DB vertically, that is, ∠A=∠ABD Let 1 be X, then ∠ DBC = 2x, ∠a =∞. ∠ACB = 90∴∠DBC+∠a+∠Abd = 90÷2x+3x+3x = 90∴∠a = 3x = 33.75 2。 CD is the bisector of ∠ACB ∠ FDC = ∠ BCD and ∠ BCD = ∠ AFD ∠ FDC = ∠ AFD, that is, △CFD is an isosceles triangle ∴ BE = EO. R congruent symmetry 80 or 20 4: 1 origin symmetry 3, 1. ∠∠ABC = 40,db = ba∴∠d =∠bad = 1/2∞。 ∠ACB = 70∴∠BAC = 1 & lt-∠ABC-∠ACB = 70∴∠△PEF =∠bad+∠BAC+∞。 X<5 1 30 III. 1. ∫ AB = AC ∴∠ B = ∠ C: Ad = AE ∴ BD = Ce: M is the midpoint of BC∴BM = cm∶: There are also ∠ ade = ∠ GDE ∴ ∠ EDF = 45 I, CBDC II, countless 15cm 65 36 III, 1. ∵AB=AC ∴△ABC is an isosceles triangle ∵ Germany ∴ BC ∴ B+∠ E = ∠ C+∠ CFD = 90 ∴∞. 0 1 2 a√b 3, 1。 ∵(2-a)? +√(a? +b+c)+|c+8|=0 ∴a=2 b=4 c=-8 ∵ax? +bx+c=0 ∴2x? +4x-8=0 2x? +4x=8 x? +2x=4 ∴x? +2x+ 1 = 4+ 1 = 52。 (1) (2) (3) The answers are all1(4) n-n+1=1i. Cadbd II and III won't. √ a 1/81/45/2 √ 59.1-0.196 III.1. =-52.= 3/23.= 9/44.= 66.= - 128 (x- 1)? = -64 x= -3 five, ∫? √= 3 ∴x=27√z-4 = 0 ∴z=4 ∴(2y- 12+2)? =0 2y= 10 y=5? √27+ 125+64= 12 2.(? √0. 125÷8)X6 = 0. 125 X6 = 1.5cm? 1. DBCB II. S=x(5+x/2)=5x+x? /2xy25 12.5s = 3n+ 13, 1.2a+ 1 = 02 b-4 = 02A =- 12B = 4A =-0.5B = 22。 (65438+.(2)0≤t≤30 (3) slightly one, BCBA two, y = 1/3x+43 The third space will not = = three, 1. ( 1) m = 2 (2) m = 22。 (65438+) 4 >4 x=2,y= 1 3, 1。 ( 1)x = 4(2)x < 4(3)x & gt; 4 x & lt; 5 2.( 1)y 1 = 50+0.4 x2 = 0.5x(2)50+0.4x = 0.5 xx = 500(3)y 1- 1。 The fourth answer of CCC is incorrect. Option 2, -2x minus 1-2-. X< 180 three. 1.∫ The image of the proportional function y=kx passes through a (k, 2k) ∴2k=k? K=0 or 2 ∴ When k=2, the coordinate of point B is (4,0). Combine A (2 2,4 4) B (4 4,0) and 1/2 to get y=-2x+8 (what? Link 1/2? When k=0, AACA line does not exist ∴ k = 2 y =-2x+8 2. ( 1)y = 10-2x(3≤x≤4)(2) 10-2x = 3.5。 X≤3) 2.2+ 1. 1x (in this case, a system of binary linear equations) 2.2 9.9 III, 1. Not 2. y = 600-10x (0 ≤ x ≤ 6) 3. (1) (binary linear equations here) (2)30KG I, ADB II, H = 1 200-150t (0 ≤ t ≤ 8) 2n+1iii,1. Y = 2.5x (x ≥ 65438 √ 2) ∴ 2 = k (-√ 2) ∴ k =-1∴ The proportional function is y =-x: point A(a,-4) ∴-4 =-a. B) ∴ b =-(-2 √ 2) = 2 √ 2 3。 (1) y = (x-2000) X5% = y = 0.05x-100 (2) 5 yuan (3) 0.05x. BDDA II, the 6th power of -3x, the 4th power of Y (= = cannot be typed) 8/kloc. ) to the fifth power -9x? The sixth power of y (this is a multiplication symbol) (xy? )? =-(3xy? ) to the fifth power-the fifth power of ——9x and the fifteenth power of y 2. = 10a? b? +ba? b? 3.x? -x+ 15=x? +5x+4-6x = 4-15x =11/64.5 (a power of 2, b power of 2) = C5. [(4n+3m)+(2n-3m)] (m+2n)/2 = 6n。 )÷2 =3mn+6n? 1. CCCD II. b? - 144 x? - 16y? x? + 16x+60 4 38 5 -24 2x? -Really? +2xy a? +2ab+b? -c? Third, 1. =(x? -9)(x? +9) =x? -8 1 2.= x? -(2y-3/2)? =x? -4y? +9/4 3.=a? -4ab+4b? +b? +2b+ 1+99 =(a-2b)? +(b+ 1)? +99 = 99 4. ∫ A-B = 2B-C = 2A+C =14 ∴ Let C = XB = X+2A = X+4 ∫ A+C =14 ∴ X+. -B? =32 5.∫a(a- 1)-(a? -b)=2 ∴a? -Ah.-Ah? +b=2 ∴b-a=2 ab-a? +b? /2=2ab-(a? +b? )/2=a? -2ab+b? /2 x (- 1)=-2 6.x? +y? +4x-6y+ 14=(x? +4x+4)+(y? -6y+9)+ 1=(x+2)? +(y-3)? + 1∶(x+2)? ≥0 (y-3)? ≥0 ∴x? +y? +4x-6y+ 14 & gt; 0 one, ACD two,-10a? (a+3b)(x-y) a x- 14 a? -2ab- 1( 1-a+b)( 1+a-b)(x? +5x+5)? Iii. 1.x? -Really? +x-3y =(x+y)(x-y)+x-3y = x+y+x-3y = 2(x-y)= 2 ^ 2。 (x? +2xy-3y? )⊙(xy)=(x+3y)(x-y)⊙(x-y)= x+3y = 125 3 . x+ 1+y(x+ 1)= 6(x+ 1)-b? -2bc-c? =2? -(b+c)? The sum of two sides of a triangle is greater than the third side ∴ 0 < a < b+c ∴a? & lt(b+c)? ∴a? -(b+c)? & lt0 ∴ sign is negative one, ∴××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××× The 6th power of x+the 4th power of 3ax-a? The fifth power of X is five, ∫| a-b+3 | = 0 ∴a-b=-3∫(2a+b)? =0 ∴2a+b=0 ∴a-b=3 2a+b=0 (here is a system of binary linear equations) and a= 1 b=-2 2a? b(2ab+ 1)-a? (-2ab)? =4a? b? +2a? b-4a? b? =2a? B will be 2a? B substitution evaluation, get 2x 1? X(-2)=-4 Six, (1)-2ab+BC+8ac-ab+2bc-3ac =-3ab+4bc-5ac (2)-3ab+4bc-5ac-ab-2bc+3ac =-4ab+7bc-63.