The math problems in the second year of junior high school are grateful to the little brother who ca

1.The parallelograms are bisected diagonally from each other, i.e., ao=ocbo=boaboperimeter=ao+bo+ab

BCO circumference = BC+OB+OC

i.e. ab-bc=8

ab+bc=30

The binary equation is solved, ab=19, bc=11

2.If the question is correct, the solution is as follows.

de=ec, aed= cef, ade= dcfaed cef, cb=ad, i.e., ae ad de, cf ef ce

Prove; CEF 90°, i.e. f≠90°

Let cef=a°= bae

Hence baf f

Question 3. Sorry.

1：y=x+8

60=2x+2y

2x+2y=60

2x+2(x+8)=60

x=11y=19

Question 2: You typed the wrong question, depressed, can't be equal, question 3, I'm sorry, I can't see the picture.

1 Let the long side ab be x and the short side bc be y, then x+y=30;The circumference of the triangle AOB coin is more than 8cm, AO and OC are equal, and OB and OB are equal, so 8cm is worse than X and Y, fixed X-Y=8, and the solution is X=19, Y=11.

Question 1: Original = Question 2:

Answer: x=5, y=2. Because they are two children and they are ages, x and y are positive integers, and the values are substituted from 1 to know that x=5 and y=2. Question 3:

Answer=-20052005. Original = 2003 20052005 2005 10001 2004 = 2003 20052005 20052005 2004 = (2003-2004) 20052005=-20052005. If you don't understand, you can ask me, I hope to adopt.

1)be+df=ef.

The conclusion of the amusement in 2)(1) still holds.

Proof: Extend CB to M so that BM=DFEven Gao Bi continued to am

AD parallel BC, then ABM= BAD=120°;

In trapezoidal ABCD, ab=cd, then the trapezoid is isosceles trapezoidal, so d= bad=120°= abm

abm≌⊿adf(sas),am=af; ∠bam=∠daf.

Therefore maf= bad=120°; If you sell eaf=60°, then mae= fae; ae=ae.

mae fae(sas),me=ef, i.e. be+bm=be+df=ef.

Intercept od=ob on OA

Because p is one point on the aob bisector.

So 1= 2

OP is a public line.

So dop is all equal to bop

So ob=od

obp=∠odp

obp+∠odp=180°

So oap= odp

So ap=dp

So APD is an isosceles triangle.

PC OA is the height of an isosceles triangle.

So ac=dc

oc=od+dc=oa-ac

So od+oa=2co

That is, ob+oa=2co

If the walking speed is x km/h, the speed of a bicycle is x+8 km/h, and the speed of a car is x+16,000 knaynam-h.

2 x+10 (x+16)=12 (x+8) multiplied by x(x+16) (x+8) on both sides at the same time

Get. 2 (x+16) (x+8) + 10x (x+8) = 12x (x+16) 2x potato sis +48x + 256 + 10x +80x = 12x +192x48x + 80x - 192x = -256

64x=256

x = 4 walking speed = 4 kilometers and hours of moderation.

Bicycle speed = 4 + 8 = 12 km/h.

Car speed = 4 + 16 = 20 km h.

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

3x*2x=300

6x^2=300

x^2=300/6

x^2=50

x=√50x=√2*√25

x=5 2, then length=3x=3*5 2=15 2

Width = 2x = 2 * 5 2 = 10 2

is the root number, is the power of the base on the left, and exponential on the right.

The process flow of the calculation process is impeccable.

But the answer is wrong.

Once the calculations are done, remember to verify.

There is no solution to this problem, and you can't cut out a rectangle with such a large area along the edge"400 square centimeters of square pieces of paper"

The side length is 20, which means that the length and width of the rectangle you cut must meet 20

And according to the result of that calculation, length = 15 2 >20 no solution.

And even if the length is set to 20, then the width is 40 3, and the final area of 267<300 is not up to standard.

You can't figure it out.

But you have to be clear about whether the reason why you can't figure it out is because you don't know how to calculate, but just metaphysically think that the problem is generally solved, so you get tangled, and you have to calculate yourself for a long time.

You're only in your second year of junior high school, so make sure you're foolproof.

When it comes to the college entrance examination, if there is no verification result or no verification process is written, it is likely to cause mistakes and fall short.

Generally, like yours, this question has a full score of 5 points, and if you don't verify it, you will be given a maximum of 2 hard points.

Root 2 is 15 times longer and root 2 is 10 times wider

x n (0 n 4, n is a natural number).

y n (0 n 4, n is a natural number).

For the meaning of "contained in" in mathematics, if the corresponding symbol is not found, this symbol is used instead.

a x-2 +3=1-x 2-x denominator is due to the omission of the term 3, and the solution of the equation is 5

a/(5-2)=(1-5)/(2-5)

a/3=-4/(-3)

a=44/(x-2)+3=(1-x)/(2-x)4/(x-2)+3=(x-1)/(x-2)4+3x-6=x-1

3x-x=-1-4+6

2x=1∴x=

That is, the solution of a+3=1-x is x=5

substitution, a=-7

Just remove this.

3. Bring five to x to count. Finally, calculate a.

1) 1= 2 we know ed=ec, and ae=bc, so the two rts are congruent.

2) two congruence can be known, bec= ade; From the inner angle and 180 degrees of the triangle before the stuffiness, it can be seen that ADE+ AED=90°=BEC+ AED; The flat angle is 180°, so ced=90°, i.e. ced is rt

3) CED is an isosceles right triangle, and the length of CE is easily derived from the Pythagorean theorem (the length of the hypotenuse multiplied by the cover of the root ant by 2 divided by 2).

1) Shouting Danyou Zheng Pin 1 = Chi Zao 2

de=ec and a= b

ae=bcrt ade and rt bec congruence (hl) is too annoying. I won't answer.