Two math problems first grade are urgent

1.Because a-b=3 5, b-c=3 5, the addition of the two formulas gives a-c=6 5, and the above 3 formulas are added by square each other.

a -2ab + b + b -2bc + c + a -2ac + c = 9 25 + 9 25 + 36 25, that is, a + b + c -ab-bc-ac = 27 25, and a + b + c = 1, so ab + bc + ca = -2 25.

2.Since 2m -2mn+n +4m+4=0, (m-n) +m+2) =0, so m=n=-2, substituted m n+mn =-16.

Because a-b=b-c=3 5;The sum of the two equations gives a-c=6 5;

a-b) 2=a 2-2ab+b 2=9 25b-c) 2=b 2-2bc+c 2=9 25a-c) 2=a 2-2ac+c 2=36 25 The above 3 formulas are added: 2a 2+2b 2+2c 2-2ab-2bc-2ac=54 25

Both sides of the above equation are dealt with 2 at the same time to obtain: a 2 + b 2 + c 2-ab-bc-ac = 27 25

And because a + b + c = 1

So 1-ab-bc-ac = 27 25

So ab+bc+ca=-2 25

Because 2m -2mn+n +4m+4=(m -2mn+n)+m +4m+4)=(m-n) +m+2) =0

So m=-2, n=m=-2

So m n + mn = -16

1.(1-1/2^2)(1-1/3^2)(1-1/4^2)..1-1/9^2)(1-1/10^2)

The middle covenant is about to be dropped.

2.Set $x per payment.

x=(8224-x)*(1+,x=4224.

$4,224 per payment.

1. The side length of the square is 480 4 = 120 meters, and the area is 120x120 = 14400 square meters.

The circumference of the circle is 480 = 2 r, and the area = rr = 240x (480 2) = 2880000 157 square meters.

Therefore, the area of the circle is large, 2880000 157-14400=619200 157 3944 square meters.

2. The side length of the large square is 30, the side length of the small square is 3, and the side length of the shaded part is 30-2 3=m.

1. Enclose a circle with a large area.

Enclosed in a circle: that is, the circumference of the circle is 470m, and the radius r=235 can be obtained, so the area is about;

If it is enclosed into a square: that is, the circumference of the square is 470m, and the side length can be obtained, so the area is about more.

Garden 18343 Fang 14400 How much is it to subtract yourself Brother The second figure is not.

Enclose a large area with a circle:

s=2πr r=s/2π=

s round r2=18328

S-square 14400

S difference s round - s square 3938

3x + c = 3y + c

Subtract c from both sides at the same time to get:

3x = 3y

Divide both sides by 3 at the same time to get :

x = yx - 3 = one part of x.

Both sides x at the same time get:

x² -3x = 1

Add 2 to both sides at the same time to get :

x² -3x + 2 = 3

Solution: (a x c 3y c

Subtract c 3x 3y c c on both sides to get: 3x 3y

Divide both sides by 3 to get: x y (two), x 3 1 x

x²－3x＝1

x²－3x＋2＝1＋2＝3

1. Because c is a constant, you can subtract c on both sides of the equation at the same time, and divide 3 on both sides of the equation to get x=y

2. x-3 = one-half of x. Both sides of the equation are multiplied by x at the same time

Get x squared - 3x = 1

So the square of x - 3x+2=1+2=3

Centimeter.

2. Add me, I'll give it to you another day.

1. BC = 5 or 1, trapped as B may be outside AD or between AD

1、(x+1)^5/(-1-x)^4=(x+1)^5/(1+x)^4=1+x

So 1+x=3x+7

So x=-3

2. Set the dividing formula to y, x 2+2x-1=yx-1

So y=[(x 2+2x-1)+1] x=[x 2+2x] x=x+2

The first question is x=—3

The second question is x+2

1.In fact, the function of the mirror is to find the axisymmetric figure of the letters and numbers on the rectangular strip of paper, that is, to put the word.

Whether the mother or number is still in its original shape after flipping it up and down, so the answer is obvious: b c d e 1 3

2.Cut an isosceles triangle from the middle (the top corner sticks the long side, the bottom corner sticks the short side), and then there are 5 shapes of different shapes on each side, 5+4+3+2+1=15.

Answer: 15.

1、cde3

2. Fold into an accordion shape of 3 cm by 5 cm, a total of: 30 3 = 10 pieces.

Total creases: 10-1 = 9 folded into an accordion shape of 3 cm x 5 cm, total: 30 3 = 10 pieces.

A total of creases: 10-1=9.

Solution: 1. Set the money to be how much x, then.

x*3*x=8000

2. Let the first batch of purchases be x, and the second batch of purchases be 2x, then [(x+2x)*(k 3)-16x 3-21*(2x) 4] [16x 3+21*(2x) 4]=20%.

18k-32-42)/(32+42)=

k=74/15