A sophomore math problem, urgent! Seek help!

Set team A as x people and team B as y people, and transfer z people from team B to team A.

According to the title, there is: (x-70) 2=y+70

y-z）×3=x+z

Simultaneous equations, subtract y to get:

x=126+4z/5

Because x is the number of people, if you want to take an integer, then z takes the smallest non-zero integer as 5, that is, when z=5, x=126+4=130 (people).

A: Team A has at least 130 players.

Set up X people in team A and y people in team B.

Equation 1: 2(70-x) = y+70

Equation 2: x+a=3(y-a).

Obtained by Eq. 1: y=70-2x Brought in Eq. 2:

Equation 3: x=30-(4 7)a

Because x,a must be an integer and ask team A at least, then:

x is at least 4, at which point a=9

Solution: If team A originally had x people, then team B originally had (2x-210) people; Y people were transferred from team B to team A, that is, there were:

x+y=3(2x-210-y)

Solution: x=126+4y 5

x and y are both positive integers, and x is a primary function of y: when y takes the minimum value, x has a minimum value.

i.e., when y=5, x=130

A: Team A has at least 130 players.

Set: A x; Y has set "several" to a

y+70=2(x-70) (1)

3(y—a)=x+a (2)

From (1) (2): 5 3x=4 3a+210 Because a>=0, 5 3x>=210

Get x>=126

Let bh be x, then hc=1000-x, according to the title, there is: ah=xac=2x (the right-angled side corresponding to 30° in the right-angled triangle is equal to half of the hypotenuse), in the triangle ahc:

x 2 + (1000-x) 2 = 4x 2 to solve the equation.

Author: nanrouzhen |Level 2 | 2010-11-24 21:55

Answer: BM+DN=MN

Proof: With point A as the rotation center, ABM is rotated 90° clockwise so that point B coincides with point D, and the corresponding point of M of ABM is recorded as M, and it is easy to know that M, D, and N are in the same line, and M D=MB, AM=AM, and M AD= MAB

It is known from the properties of isosceles right triangles: man=45° bam+ dan= bad- man=45° and m an= m ad+ dan= bam+ dan=45° m an= man

In AMN vs. AMN.

am=am′

an=an∠m′an=∠man

am′n≌△amn

m′n=mn

m′d+dn=mn

mb+dn=mn

a²-b²)(a²+b²-c²）=0

a -b = 0, or: a + b -c = 0

a=b (a=-b is not appropriate, discard), or: a +b =c The triangle abc is an isosceles triangle or a right-angled triangle.

It should be missing 1 solution.

In one case, when a=b, the square difference between a and b cannot be approximated. All answers should be right triangles or isosceles triangles.

A -b may be equal to 0, so it cannot be divided directly. When a = b, it is an isosceles triangle; When a<>b is a right triangle.

There is no discussion, ab equal cases. If ab is equal, it is also possible. Isosceles triangle.

The quadrilateral MNCO is a parallelogram.

mn=oc by the title, angle a'cd=right angle.

The hypotenuse of a right triangle is equal to the hypotenuse of the general.

oc=1/2da'=（bc－ad）÷2=mn

Translate DC and MN so that AMD coincides, and it can be seen that N' is at the midpoint of BC', and MN (BC AD) 2 can be deduced from the rectangular diagonal theorem

Can you not use panning??? Is it okay to make auxiliary lines???

You may wish to set Xiao Ming to buy a kilogram of eggs every time, Xiao Gang to buy B yuan of eggs each time Xiao Ming two purchases total** for ax+ay, a total of 2a kilograms of Xiao Ming two purchases average ** for (ax+ay) (2a) = (x+y) 2 yuan kg.

Xiao Gang's two purchases totaled ** 2b, and a total of b x + b y kg Xiao Gang's two purchases averaged ** 2b (b x + b y) = 2 (1 x + 1 y) = 2xy (x+y) yuan kg.

Compare the average unit price (x+y) 2 and 2xy (x+y) for two people

x+y)/2 - 2xy/(x+y)

(x+y) 2 - 4xy] [(x+y) 2]= [(x-y) 2] [(x+y) 2] Since x is not equal to y, and x,y are both greater than zero.

So [(x-y) 2] [(x+y) 2] >0 so (x+y) 2 - 2xy (x+y) >0 Xiaogang's way is cost-effective.