A comprehensive math problem for the second year of junior high school, God,,, give me a method toni

Little classmate, you have a problem with the question, take a closer look to see if there is a letter typed wrong? Or send a ** over and I'll help you answer.

This seems to be a wrong question, since the triangle is an isosceles right triangle, you might as well take the point d as the midpoint of BC, then C and F coincide, AD obviously cannot be equal to BC, that is, BF

You can first calculate how much is 1 kilogram of bananas, how much is 1 kilogram of fresh oranges, 1 kilogram of bananas = m a yuan.

1 kg of fresh orange = m b yuan.

3 kg of bananas and 2 kg of fresh orange = 3m a + 2m b

Bananas per kilogram **m a

Fresh Cheng per kilogram **m b

So 3m a + 2m b

Answer: 26 or 28

Solution: ABCD is a parallelogram.

ad∥bc∠aeb=∠cbe

AE bisects ABC

abe=∠cbe

i.e. aeb= abe

ae = ab due to ad = ae + de = 9

When ae=4, then ab=4

So the circumference of the parallelogram ABCD = 2 (ab + AD) = 26 and when AE = 5, then ab = 5

So the circumference of the parallelogram ABCD = 2 (ab + AD) = 28 To sum up, the circumference of the parallelogram ABCD is 26 cm or 28 cm

The title is wrong, why are there parallel lines in the corners?

Square side length = 20 cm.

The length of the rectangle is 3x, and the width is 2x

3x)*(2x)=300

x^2=50

x = 5 root number 2

So length = 15 root number 2

Because (15 roots, number, 2) 2 = 450> Qi sedan nucleus, 20 2, so high digging 15 roots, number 2, 20

Therefore, the required rectangle cannot be cut.

The square is 20*20

Let the length of the rectangle be x and the width be y, then xy=300

x/y=3/2

The solution is 450 for x squared, and x must be greater than 20

It was already longer than a square piece of paper, so he couldn't cut a piece of paper that met the requirements for a lack of rubber ridges.

Solution: If the plane should return after flying a maximum of x kilometers, then there is equation:

x/(800+20)+x/(800-20)=9x/820+x/780=9

1/820+1/780)x=9

x≈3598

A: The aircraft should fly up to 3,598 kilometers before returning.

The aircraft should return after flying a maximum of xkm.

x/(800+20)+x/(800-20)≤9x/820+x/780≤9

780x+820x≤9*780*820

x km. The aircraft should return after a maximum of 3,598 kilometers.

Set to fly out x km.

then x (800 + 20) + x (800-20) = 9x (1 820 + 1 780) = 9

x = A: Up to 3598 km.

Let the return time be t

800+20）t=(800-20)(9-t)820t+780t-7020=0

1600t-7020=0

t= can fly up to 820t=

1. 1+3+5+7+..29 = 225 = (1 + 29) * 15 2 = 225 [the sum of the first 15 terms of the equal difference series] 2

5+31)*(16-2) 2=252 [sum of the first 16 terms of the equal difference series] 3 1+3+5+..2003=522

1+2003)*1002 2=1002 2=1004004 [the sum of the first 1002 terms of the difference series].

ab=1a^2+b^2

a+b)^2-2ab

a-b)^3 (a-b)^2(b-a)

a-b)^3 (a-b)^2[-(a-b)]=-(a-b)^(3+2+1）

(a-b)^6

The purpose of the first question is to let you find the pattern, so don't use the summation formula, 225 is 15 squared.

252 is 16 squared minus 2 squared.

Patterns can be derived.

1+3+5+……2n+1=(n+1) 2So1+3+5+.2003=1002 22, this relies on the use of a completely flat way.

a²+b²=(a+b)^2-2ab=9-2=73、（a-b）³（a-b)²（b-a)

(a-b)^3*(a-b)^2*(a-b)=-(a-b)^(3+2+1)

(a-b)^6

When multiplying, the bases are the same, the exponents are added, and the symbols are noted.

1.No, it won't.

2.(a+b)²=a²+b²+2ab

9=a²+b²+2

a²+b²=7

3.No, it won't.

PS: Just do what I can

1】(1+2003)×1002÷2＝104004

2] Original (a b) 2ab 3 1 2 7

3] The sixth power of the original formula (a b).

2. Using the perfect square formula, the final is equal to 1

: Set up a first-class car X section, a second-class car Y section.

64x+92y=496，x+y=6

The solution is x=2 y=4

A: There are 2 cars in 1st class and 4 cars in 2nd class.