8th grade math problems, please write the process clearly, thank you

If the side length is 4 and the ap is 1, you will know that the p point is on ab.

There are three ways to find it:

1. Projective theorem (I guess you haven't learned it, so I'll talk about the second method).

2. In the RT triangle BPC, BP 4-1 3, BC 4, so PC 5 (Pythagorean theorem can be found), let PE X, then CE 5 X, then the triangle BPE and triangle BEC are RT triangles, so the Pythagorean theorem can be obtained: the square of be 3 squares - the square of X 4 (5 X) squared.

Solving an equation can find x, so the square of be 3 and then the root number can be found, and the answer is be.

3. Area method. Otherwise, this method is the simplest: the area of the triangle BEC is 4*3 divided by 2 6

The area of the triangle BEC is 5*BE divided by 2 6 so BE

It's not difficult for you to draw a picture and think about it slowly. I don't know if there's a miscalculation, so you do the math carefully. The third method is recommended. Good luck in your studies!

No diagrams?? Then I'll assume:

1. If the P point is on the AB line, then BP=AB-AP=3, BC=4, then PC=5, according to the triangle comparison.

be bc=bp pc, then be=3 5*4=2, p point is on the ad line, then be=ab=4

1. If the P point is on the AD, then:

Because CBE+ BCE= PCD+ BCE=90 degrees.

So cbe= pcd

In right-angle CPD and BCE, because CBE= PCD, BEC= PDC (right angle).

So CPD is similar to BCE

Therefore, be cd=bc cp, be=(bc cp)cd is known as bc=4, cd=4

And cp= (cd +pd )

cd²+（ad-ap）²）

So be=(bc cp)cd=(4 5) 4=2, if the p point is on ab, then:

In BCP and BCE, BCP = BCE (same angle) and PBC = BEC (right angle).

So BCP is similar to BCE

Therefore, be bp = bc pc, be = (bc pc) bp is known bc = 4, bp = ab-ap = 4-1 = 3 and pc = (bc + bp ) = (4 + 3 ) = 5 so be = (bc pc) bp = (4 5) 3 = 12 5 =

The integer part of 13 should be 3, i.e., m=3, then n= 13-3

m-n=6-√13

The integer of 13 is 3, so n = 13-3

So m-n=3-(13-3)=3- 13+3=- 13

So the answer is negative 13

Integer calculations.

1) 3a[b²-3a(b-3a)]+b(9a²-3ab+b²)=3a(b²-3ab+9a²)+9a²b-3ab²+b³=3ab²-9a²b+27a³+9a²b-3ab²+b³=27a³+b³

2)(4x-½y)(x²-2xy)

4x³-8x²y-½x²y+xy²

4x³-17/2x²y+xy²

There is no skill in these two questions. That's it. The main thing is to be serious!

Pay attention to the order of the letters when writing (e.g. a in front of B and x in front of Y, so that it is not easy to make mistakes).

Factorization. 1) ab-a+b-1

a(b-1)+(b-1)

b-1)(a+1)

Mention A, then B-1).

2)(a+b+c)²-a-b-c)²

a+b+c+a-b-c)[(a+b+c)-(a-b-c)]=2a x 2(b+c)

4a(b+c)

Resources. a²-b²=(a+b)(a-b)

The absolute value and the sum are greater than or equal to 0, and the addition is equal to 0, if one is greater than 0, the other is less than 0, which is not true.

So both are equal to 0

So x+y-5=0

xy-6=0

i.e. x+y=5, xy=6

So (x+y) = 5

x²+y²+2xy=25

x +y = 25-2xy = 25-12 = 13 (x-y) = x +y -2xy = 13-12 = 1, so x-y = -1 or 1

Solution: (Is C-4 whole, or is it separate?) )

Separate 4 a, 8 b, 16 c-4

2^2a÷2^3b÷2^4c-4

2^(2a-3b-4c)-4

Because 2a-3b = 4c-4

i.e. 2a-3b-4c=-4

So the original formula = 2 (-4)-4

4 A 8 B 16 (C-4) in the whole

2^2a÷2^3b÷2^(4c-16)

2^(2a-3b-4c+16)

ab=ae

Angular abe=aeb

AD parallel BC

Angular AEB = EAD = 2 BAE

ADCE is isosceles trapezoidal.

Angular EAD = CDA = 2dab

dab=bae

bf = AF angle BFE = FAB + FBA = 2 FAB = EAD = Beabe = BF

Therefore af=be

Put the oil tank on the side, is a rectangle, find the diagonal.

Root number (12 2 + 5 2) = 13m