Junior High School Math Problems!! Correct high bounty !!

The vertex p of y=ax +bx+c(b>0,c<0) is on the positive semi-axis of the x-axis.

b^2=4ac

b+ac=3

So b=2ac=1

q(0,c)

p(-1/a,0)

0a=0p*0q root number: 0p 2+0q 2

2=(c^2/a^2)/(c^2+1/a^2)2(a^2c^2+1)=c^2

c=-2a=-1/2

The parabolic equation is y=-1 2)x 2+2x-2

Calculate one first, b=2 a is equal to the root two of the negative half. c is equal to the negative root two.

From the meaning of the title, the axis of symmetry - b 2a=2

When x=0 y=c=-2

And because b+ac=3

The solution yields a= b=2 c=-2

Solution (1) is based on b+ac=3

and the vertex p is on the positive semi-axis of the x-axis (vertex ordinate = 0, giving the square of b = 4ac) so b = -6 (rounded) or b = 2

2)ac=3-2=1

Then according to the height of the hypotenuse of the right triangle OPQ is the root number 2, we get c=so y=-1 2x +2x-2

Establish. Reason: Because a b=k, a=bk, the same way c=dk

So a-c=bk-dk=k(b-d), so a-c b-d=k, the same way a+c b+d=kThat is, a-c b-d=a+c b+d is true.

Established by a b = c d = k, a c = b d

Let a c=b d=m

then a=cm, b=dm

a-c=cm-c=c(m-1)

b-d=md-d=d(m-1)

a-c b-d=c(m-1) d(m-1)=c d.

a+c/b+d=c/d

So a-c b-d=a+c b+d holds.

This is the sum ratio theorem and the difference ratio theorem.

is obtained by a b = c d = k.

a=bk c=dk

a-c/b-d=bk-dk/b-d=k

a+c/b+d=bk+dk/b+d=k

The conclusion of the topic can be obtained.

I guess I've learned the equations in junior high school.

1.If workshop A produces x pieces per day and workshop B produces y pieces per day, then y-x=3 (1).

3x-2y=1 (2)

1)*2+(2) gives x=7, substituting (1) gets y=10

2.Set up x pieces of product A and y pieces of product B.

x+y=100 (1), 160x+210y>=18500 (2), 2)-(1)*160 to get y>=50

x+y+100 (1), 160x+210y<18650 (2), 2)-(1)*160 get y<53, because it was produced for 7 days, all product A has 49 pieces, product B has 70 pieces, and all purchase plans have.

1: 49 pieces of product A and 51 pieces of product B.

2: 48 pieces of product A and 52 pieces of product B.

1. Solve the A workshop to produce x pieces every day, and workshop B to produce the Y piece column equation system y-x=3 3x-2y=1 to solve the x=7 y=102 solution to purchase X pieces of A, and the purchase of B Y pieces is solved by the question of x+y=100 column inequality 18500<160x+210y<18650 to get 50, so scheme 1: A49 B 51

Scheme 2: A 48 B 52

First question:

Solution: Set up workshop A to produce x pieces per day, and workshop B to produce Y pieces per day.

The system of column equations y x=3 3x 2y 1 is solved: x 7, y 10

The second question: Solve the purchase of X pieces of A and purchase of Y pieces of B.

Column equations x y 100 18500 160x 210y 18650

From: y 100 x, substitution gets: 18500 160x 210 (100 x) 18650 solves: 47 x 50

x＝47 ， y＝53

x＝48 ， y＝52

x＝49 ， y＝51

x＝50 ， y＝50

(1) Let A produce x pieces per day 2 (x+3) + 1 = 3x solution x = 7 Answer: A 7 pieces B10 pieces (2) Let A buy x pieces 18500<160x + 210 (100-x) < 18650 47

Proof: Make DB parallel AB , intersect BC with point B', because ad is parallel to bc, ab is parallel to db', so quadrilateral abb'd is the parallelogram ab=b'd, because ab=dc, so db'=dc.

The main owner of the building proves the results only at ABB'd can only be realized in the case of a diamond, and in the given condition, it is an arbitrary isosceles trapezoid. So this question is problematic, I hope you think about it again and be happy to answer it for you, I wish you progress in your studies!

If you don't understand, you can ask!!

Remember to adopt it!!

Proof: pass the D point as a DE parallel AB, cross the BC edge to the E point, connect AE, extend the BA and CD to the H point, connect the HE to the F intersection HE to the O point.

From the question, we can know that the trapezoidal ABCD is an isosceles trapezoidal ABC= dCE AD BC FDC= DCE

The quadrilateral abed is a parallelogram.

ab=de, and ab=dc

de=dc∠dec=∠dce

AD BC Hda= DCE ADE= DEC= DECa= DCE

hda=∠ade

In the same way, if there is had= dae, then ch ae in the triangle had and triangle eda have ah=dh ae=de quadrilateral, haed is diamond.

I think you should be able to make it at this point.

The conclusion of this question is not valid!!

The diagonal lines of an isosceles trapezoidal are perpendicular to each other and do not bisect the base angles.

Unless the top bottom = waist.

So what they can prove is not right!

You can draw an isosceles trapezoid, look at their diagonals, and obviously won't divide the bottom corners!

The condition cannot be ab=dc, because there are many cases of the diagonal of an isosceles trapezoid, and only when ab=ad can the angle be bisected. You're taking a closer look.

If ab=ad is not given, then it cannot be proven. You can use the counter-evidence to push it to know.

Personally, I think there is a problem with this question, and only ab ad can the conclusion be established.

The occurrence of the IK is sprinkled loudly Astonas

daskldjlasd

daskdl;kasjd

daskdowdasdasdkjdaspiuwoe jfmdasdasdasasdasasasddsadasasd

According to the existing conditions, it can only be withdrawn: the trapezoid is an isosceles trapezoid.

The conditions for justifying this conclusion are not sufficient.

Let x=1 x, 1 yy=1 y, then x+y=3 is reduced to x*y (2x-y), and the result of the opening of -y is y( ( (3-y) (6-3y))-1) is the open root number.