Find the first three question sets, and solve the three junior high school questions

I sent you another and received a letter. Satisfaction is awarded.

i.e. proof is required (ab+cd) =(b +d) (a +c ) while (ab+cd) -b +d (a +c ) = (a b +c d +2abcd) - (a b +a d +c d +c b )

2abcd-a²d²-c²b²

2（ad-bc）²

0 So (ab+cd) = (b +d) (a +c ) is proven.

For example, if you have three numbers x, y, z, you have to prove that y is the middle term in the ratio of x, z. Just prove y*y=x*z

For this question,,, because (b 2+d 2)*(a 2+c 2)=a 2*b 2+b 2*c 2+d 2*a 2+c 2*d 2

Because AD=BC , B 2*C 2=A*B*C*D , D 2*A 2=A*B*C*D, so (B 2+D 2)*(A 2+C 2)=A 2*B 2+2*A*B*C*D +C 2*D 2

ab+cd)^2

The original question is proven.

It's easy! Just put pen to paper and come out! First (b+d)(a+c) to get the equation a, then (ab+cd) to get the equation b, using the known ad=bc to compare the ab formula substitution to get a=b

It is known that ab + cd is the middle term of the ratio of b square + d square to a square + c square, i.e. (b square + d square) (ab + cd) = (ab + cd) (a square + c square) equation a square multiplied b square + a square multiplied d by d square + b squared multiplied c square + c squared multiplied d by d square = a square multiplied by b square + abcd + cdab + c squared multiplied d by d squared.

Because ad=bc.

A square multiplied by D square = ABCD = B square multiplied by C square.

Solution: According to having two identical roots, that is.

Discriminant =0 for root

i.e. =b -4ac=(-6) -4 1 k=0 solution k=9

The answer is no (1, 2).

Upstairs is doing a good job, so take him if you understand it.

I'd love to help you, but I haven't read the third year of junior high school.

(1) Let the side length of the small square be x, then the columnar formula: (100-2x)(50-2x)=3600

i.e. (x-70) (x-5) = 0, so x = 70 or 5, and x should be smaller than the side length, so x = 5cm

Question 1: 100*50=2*100x+2(50-2x)x+3600, which is reduced to -x 2+75x-350=0

It is obviously impossible to solve x=70 or 5, but 70 is greater than the width of the original tin box. So x is 5, and the second question means:

There are x teams, and every two teams play each other, for a total of 28 long games.

The process is just that:

On the first occasion, any team A1 was selected and played once against the remaining X-1 teams, for a total of X-1 games. And remove team A1 because A1 is no longer required to play against any team.

On the second occasion, a random team A2 was selected and played once against the remaining X-2 teams, for a total of X-2 games. And remove team A2 because A2 doesn't need to play against any team anymore.

.In the X-1 round, any team AN-1 was selected and played once against the remaining 1 team, for a total of 1 match. And we'll get rid of both teams AX-1 and AX, because AX-1 and AX don't need to play against any team anymore.

Take this x-1 time, and the number of games each time adds up to 28 games, i.e. (x-1) + (x-2) +2+1=28 to the left is the series of equal differences, ie.

x(x-1) 2=28, and x is greater than 0, so x is 8

(1) Solution: Let the side length of the small square be x

100-2x）（50-2x）=3600

2）28=（x-1）+（x-2）+…

Try until the result is equal to 28).

The deformation of the two questions will change it by itself, and bring X forward.

x , the most hated math Le. Not unconsciously opened -