Beijing education publishing house new classroom holiday life winter vacation book eighth grade mathematics, Chinese answer, ah, speed!

These questions are arranged in order from the first page.

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mathematics

I. DACB II. AC=DF 45cm

Third, 1. ∫AO share ∠BAC

∴DO=EO

∴∠DOB=∠EOC

∵CD⊥AB BE⊥AC

∴ODB=∠OEC=90

Ⅶ in △BOD and △COE

∠ODB=∠OEC

DO=EO

∠DOB=∠EOC

∴△BOD≌△COE

∴OB=OC

2.AD = CE，AC=BC，∠A=∠ACB=60

∴△ACD≌△BCE(SAS)

∴∠ACD=∠CBE

∫△ABC is an equilateral triangle

∴∠ACD+∠BCD=60

∴△BFC，∠bfc = 180-∠CBE-∠BCD = 120。

I. DDDA II. BD = B'd' 180。

(1)∫△ABC is an equilateral triangle.

∴AC=BC

∫△DCE is an equilateral triangle

∴DC=CF

Every angle of an equilateral triangle is 60 degrees.

∴∠ACB=∠DCF=60

∴∠ACB+∠ACD=∠DCF+∠ACD

That is ∠ACE=∠DCB.

Ⅶ in △ACE and △BCD

AC=BC

∠ACE=∠BCA

CF=DC

∴△ACE≌△DCB

∴AE=DB

(2) established.

I. DDC II. 30 3 cm 90

(1) △ ace △ CBD can be obtained by HL theorem.

∴AE=CD

(2)BD=CE= 1/2BC=6cm

I. CDCB II. ∠ ADC = 2 ∠ ABC 50

Three. 1. As shown in the figure, the intersection point D is perpendicular to AB and AC respectively, and the vertical feet are G and H respectively.

Then ∠ gad+∠ adg = 90, ∠ had+∠ dah = 90.

That is ∠ BAC+∠ GDH = 180.

∴∠BAC+∠EDF= 180

∴∠GDH=∠EDF

∴∠GDH-∠GDF=∠EDF

That is, ∠EDG=∠FDH.

AD is the angular bisector.

∴DG=DH

∴∠DGE=∠DHF=90

∴△DEG≌△DFH

∴DE=DF

2. Using the principle of symmetry axis

First, DCCC second, congruent symmetry axis

3cm

B(-2，2) C(-2，-2)D(2，-2)

four

70 or 55.

Three. ∵△ABE's circumference is 10.

AE+BE+AB= 10。

And in △AEC, d is the midpoint of AC, ED⊥AC.

∴ED is the middle vertical line of triangle AEC.

That is, △AEC is an isosceles triangle.

∴AE=EC

CA = CB

∴AC+AB= 10

∫AC-AB = 2

∴AC=6 AB=4

∴CA=CB=6

I. CC II, 2 8 or 6 rectangles 1, 3, 8, 0 15

Third, 1. ∫DE vertically divides AB.

∴AD=DB

That is, ∠A=∠ABD.

Let 1 be x, then ∠DBC=2X, ∠A=∠ABD=3X.

∠∠DBC+∠A+∠ Abd +∠ ACB = 180, ∠ ACB = 90.

∴∠DBC+∠A+∠ABD=90

∫2X+3X+3X = 90°

∴∠A=3X=33.75

2.∫EF‖BC, BD is the bisector of ∠ABC.

∴∠EDB=∠DBC

Once again, ∠DBC =∠ Abd

∴∠EDB=∠ABD

∴△BDE is an isosceles triangle

And \ef \bc, CD is the bisector of ∠ACB

∴∠FDC=∠BCD

And material ≈BCD =∠AFD.

∴∠FDC=∠AFD

That is, △CFD is an isosceles triangle.

∴BE=EO CF=DF BE+CF=ED+DF=EF

∴EF=BE+CF

1. DCBD II. F, g, k, n and r are congruently symmetric 80 or 20 4: 1. The origin is symmetrical.

Three. 1.∫∠ABC = 40，DB=BA

∴∠D=∠BAD= 1/2∠ABC=20

∫∠ACB = 70，CE=CA。

∴∠E=∠EAC= 1/2∠ACB=35

∠∠ABC = 40，∠ACB=70

∴∠BAC= 180 -∠ABC-∠ACB=70

∴∠DAE=∠BAD+∠BAC+∠EAC= 125

2.∫△ABC is an equilateral triangle.

∴∠ACB=∠ABC=60

PE‖AB，PF‖AC

∴∠PEF=∠ABC=60，PFE=∠ACB=60

The remaining △PEF is an equilateral triangle.

1.BADCC 2。 The median line of its two opposite sides is 0.

Third, 1. AB = AC

∴∠B=∠C

AD = AE

∴BD=CE

M is the midpoint of BC.

∴BM=CM

∴△DBM=△ECM

∴DM=EM

∴MDE is an isosceles triangle.

2. the perpendicular line passing through point d intersects EF at point g.

∠CDF=∠GDF, and ∠ADE=∠GDE.

∴∠EDF=45

1. CBDC II. Countless 15cm 65 36

Third, 1. AB = AC

△ ABC is an isosceles triangle.

∵DE⊥BC

∴∠B+∠E=∠C+∠CFD=90

∴∠E=∠CFD

∠∠CFD =∠AFG again

∴∠E=∠AFG

△ AFE is an isosceles triangle.

The midline of the base of an isosceles triangle bisects the base vertically.

∴AG‖BC

leave out

Tip: √ = root sign

First, CDCAB II. x & gt0 1 2 a√b

Third, 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=5

2.( 1)(2)(3) The answers are all1(4) n-n+1=1.

I. CADBD

Two, three won't = =

First, DD second, cube root? √a 1/8 1/4 5/2√5 9. 1 1-0. 196

Three. 1.=-5 2.=3/2 3.=9/4 4.=60

Four. 1.2x- 1 =√5 2x =√5+ 1x =(√5+ 1)/2

2.2(x- 1)？ = - 128 (x- 1)？ = -64 x= -3

Verb (abbreviation of verb) ∵? √=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？ /2 x x y 25 12.5 S=3n+ 1

Three. 1.2A+ 1 = 02 B- 4 = 02A =- 12B = 4A =-0.5B = 2。

2.( 1)Q=200+ 10t (0≤t≤30)

(2)0≤t≤30

(3) Omission

1.BCBA 2。 Y= 1/3x+4 3 The third space will not be = =

Three. 1.( 1) m = 2 (2) m = 22。 (1) y = 50-2.5x (0 ≤ x ≤ 20) (2) omitted.

I. CBB II. Y = X-2 3

Three. 1.( 1) x = 4 (2) x

2.( 1)y 1 = 50+0.4x y2 = 0.5x(2)50+0.4x = 0.5x x = 500(3)y 1

First, CCC's fourth answer has no correct option.

Two. -2x reduction1-2-1-5y =180-1/2x0

Third, the image of 1 ∫ 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, the AO line does not exist.

∴k=2 y=-2x+8

2.( 1)y= 10-2x (3≤x≤4)

(2) 10-2x=3 x=3.5

I. AACA II. 1200- 150 t0≤t≤845y = 2.2(0

Three. 1. is not 2.y=600- 100x(0≤x≤6).

3. (1) y = 0 (0 ≤ x ≤ 30) 0.2x-6 (x ≥ 30) (here, it is a binary linear system).

(2)30 kg

I. ADB II. h = 1200- 150t(0≤t≤8)2n+ 1。

3.1.y = 2.5x (x ≥10) y is a linear function of x and a proportional function.

2.∫ function y=kx passing point (√2, √2)

∴√2=k(-√2)

∴k=- 1

The positive proportional function is y =-x.

Point A(a, -4) is on the image.

∴-4=-a

∴a=4

∫ point B(-2√2, b) is on the image.

∴B= -(-2√2)=2√2

3.( 1)y =(x-2000)X5 % = y = 0.05 x- 100

②5 yuan ③ 0.05x-100 =1920.05x =101.92x = 2038.4.

First, BDDA II, the sixth power of -3x, the fourth power of Y (= = can't be typed), the twelfth power of 8 1a, 19 3 1, and the fourth small question will not be 4 -2.

Third, 1. = -(3xy？ ) 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- 15 x = 1 1/6

4.5 (a power of 2 and b power of 2) = c.

5.[(4n+3m)+(2n-3m)] (m+2n)/2

=6n(m+2n)

=(6mn+ 12n？ )÷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