Mathematics in the eighth grade, mathematics in the eighth grade

Set to x, then 4 x+

x=20

Hope it helps.

If you buy x items for the first time, the original unit price of each item is 4 x; Because the second time I bought a commodity, it must be greater than 12, otherwise the price cannot be reduced, and it cannot be bought more than the first time, and buying more than a dozen can be bought according to the ** after the price reduction, so it is equivalent to all the goods are bought after the price reduction, therefore, the unit price after the price reduction is 4 x - 4 x - 4, and the solution is x=20

Verify, the original unit price is 4 20 = yuan, a dozen yuan, after the price reduction, a dozen yuan, each piece, 4 yuan can buy 4 * 12 pieces.

Hehe, it's so hard.

Let the original ** of this small commodity be X yuan, and the second ** if you buy Y pieces, it will be (yuan, the number of pieces. According to the meaning of the question, the system of equations can be obtained:

xy=solution:x=

y=20

If you buy x pieces for the first time, it will be 4 x yuan per piece, and if you buy x pieces for the second time, you will have 4 yuan per piece (4 yuan, and the equation will be listed according to the problem.

4 solution gives x=20

This method is easy to understand.

Solution:1(1) a=30 degrees ad=3 2

2) a=60 degrees ad=9 4

2.The quadrilateral EDBC is diamond-shaped.

Due to the angle EOC = angle AOD

AO = OC (because point O is the midpoint of AC).

Angular ECO = Angular OAD (because EC is parallel to AB).

So the triangle EOC is congruent with the triangle DOA.

Hence ec=AD

When the angle A=90 degrees, the triangle AOD is similar to the triangle ACB.

Because point O is the midpoint of AC, then point D is the midpoint of AB, and there is AD=DB because in the triangle ABC, the angle C is equal to 90 degrees, the angle B is equal to 60 degrees, and BC=3 so AB=6, AD=DB=3

That is, ec=ad=db=3

Because the EC is parallel and equal to DB, the quadrilateral EDBC is a parallelogram.

And because in the parallelogram EDBC, ec=ad=db, the parallelogram EDBC is diamond-shaped.

1) 30 degrees, AD = 63 at the root of 2

2) 60 degrees, AD=3

3) It is diamond-shaped

If you can't see clearly, look here.

(1) According to the above law, there are a=2x1=12, b=x2, 2=5;

2) The nth equation is 2*(n+2) x-1 (x-(n+1))=1;

Solution: x1=n+2;

x2=2*（n+1）；

There are few questions and conditions, so I guess there are fewer questions.

In trapezoidal ABCD, AD BC, AB=DC, points E, F, and G are on the edges AB, BC, CD, respectively, and AE=GF=GC

1) Verify that the quadrilateral AEFG is a parallelogram; (2) When fgc=2 efb, it is verified that the quadrilateral AEFG is rectangular

Proof of: (1) In trapezoidal ABCD, ab=dc b=c

gf=gc∠c=∠gfc

b=∠gfc．

AB GF, i.e. AE GF

and ae=gf

The quadrilateral AEFG is a parallelogram

2) Pass the point G as GH FC, and the vertical foot is H

and gf=gc

fgh=1/2∠fgc．

fgc=2∠efb

fgh=∠efb．

fgh gfh=90°

efb＋∠gfh=90°

efg=90°．

The quadrilateral AEFG is a parallelogram

Quadrilateral AEFG is rectangular

Here's what we're talking about. As shown in the figure, in the trapezoidal ABCD, AD BC, AB=DC, points E, F, and G are on the edge AB, BC, CD, respectively, and AE=GF=GC

1) Verify that the quadrilateral AEFG is a parallelogram; (2) When fgc=2 efb, it is verified that the quadrilateral AEFG is rectangular

Proof: (1) Prove that in trapezoidal ABCD, ab=dc b= c

gf=gc∠c=∠gfc

b=∠gfc．

AB GF, i.e. AE GF

and ae=gf

The quadrilateral AEFG is a parallelogram

2) Solve the point g as gh fc, and the vertical foot is h

and gf=gc

fgh=1/2∠fgc．

fgc=2∠efb

fgh=∠efb．

fgh gfh=90°

efb＋∠gfh=90°

efg=90°．

The quadrilateral AEFG is a parallelogram

Quadrilateral AEFG is rectangular

Solution: Root number (k-3 2k-3) = root number k-3 root number 2k-1k-3 2k-3 = k-3 2k-1

2k²-9k+9=2k²-7k+3

9k+9=-7k+3

9k+7k=3-9

2k=-6k=3Answer: The condition for this equation to hold is k=3

k must be greater than or equal to 3 for the equation to make sense.

The denominator on both sides of the equation is unequal, and if the numerators are to be equal, and the equation is guaranteed to hold, then the numerator can only be 0

then k-3=0 and k=3

Because in the parallelogram ABCD dab=120°, so d=60°, and because ah cd,ad=2, then dh=1, ah= 3, i.e. the distance from point A to Cd is 3.

Because the quadrilateral CEFG is folded from the quadrilateral AEFD, and the point A is folded and coincides with the point C, so ADF CGF, that is, the area of ADF and CGF is equal, and AC and EF are bisected perpendicular to each other, we can know that the quadrilateral AECF is a diamond, there are AF=CF, AFC= AEC, so AFD= BEC, because CD=3, DH=1, so CH=CF+FH=AF+FH=2, let FH=X, then AF=CF=2-X, at right angles In AHF, the Pythagorean theorem has ah +fh = af , that is, ( 3) +x = (2-x) , the solution is x=1 4, so df = 5 4, from d = b, afd = bec, ad=bc afd bec(aas), so df = be=5 4, cgf area = adf area = df ah 2=(5 4) (3) 2=(5 3) 8.