Problems with the speed of solving problems in junior high school math!!

I'm in college now, I remember when I was in junior high school, I studied mathematics, especially proof questions, and I was more distressed, but in the end, I think that no matter how difficult it is, the topic must be the knowledge points we have learned, and the main reason why we can't do it is that we haven't practiced it yet, that is, when I see such problems, I don't even have ideas! So I was starting from scratch, I had to understand every question the teacher said, one way was to ask the teacher, and then use a special book to write down the steps according to their own ideas, straighten it out, and I had to re-ponder the questions I had memorized, so as to find where the teacher's thinking about this question begins! And how did you find the answer to your question step by step?

I'm also a chronic student, but although I do less, I can guarantee that I do all the questions correctly, thinking that blindly pursuing fast, the quality must not be up to par! The teacher who produced the paper must not be able to answer the paper by ordinary students, because there are still more powerful classmates Or everyone has a high score, right?

Anyway, I'm just going step by step, and among the many factors in doing the question, the first thing I consider is whether I can get it right after I finish answering! If it's not right, it's all in vain. Then increase the speed on this basis, you have to ask the way to increase the speed, I can only say that the speed is what you have after doing a certain amount of questions!

It's not a sea of questions tactic, but after doing some questions, look back and think about what these proof questions have in common and what ideas have in common! After summarizing, if you do the question again, you will feel that you have an idea, and your speed will naturally increase accordingly.

You tell me? I'm not entirely right, I'm stupid, but for me, I did see hope that you can learn from it. o(∩_o~~~

Calm down, don't be scared, only do 5 out of 10, but make sure that all 5 are right, I think you can't reach it if you want to hurry.

Read some questions every night and remember the ideas! Then repeat the idea every day, so that you don't have to read a lot of questions every day! Then remember the train of thought and review him, which should add up to a lot, insist on it every day, read the questions every day, and there are more than 1,000 questions in a semester!

Are you still afraid of problems? That's what I did, and it worked very well, focusing on the problems I didn't know!

According to the vertical diameter theorem, it can be known that AM=3, connected OA, and the Pythagorean theorem OA=5

7.From the meaning of the title, we can know that dh = root number 2, the Pythagorean theorem can be found bh=1, and then the intersecting string theorem, chxdh=ahxbh

Bh=2, so ab=2+1=3

11. According to the analysis of the title, the diameter is the longest string in the circle, so it is the diameter.

af=cd, angle a=angle e, ab=de These 3 conditions deduce abf congruence dec bfa= ecd deduce bfc= fce deduce bf ce and is parallel and equal by congruent bf=ce so bfec is a parallelogram 2 q: bcef is a diamond then bc=bf Since b=90° from b point to AC side perpendicular to m bcf in acb According to the question set, it can be seen that cm=9 5 fm=cm then af=5-2*9 5=7 5 classmate Save your life and take it You know.

1. ABF CDE (as far as known conditions are known).

bf=ce，∠afb=∠dce

ac=af+fc，df=fc+cd

ac=df△abc≌△def

BC=EF, so the quadrilateral BCEF is a parallelogram.

2、ac=√(3²+4²)=5

If the BCEF is diamond-shaped, then Fc and Be are bisected perpendicular to each other, so that the intersection of the bisector line is O

rt△bco∽rt△abc

oc/bc=bc/ab

oc=bc²/ab=9/5=

af=ac-2oc=

When af=, the quadrilateral BCEF is rhomboid.

Supplementary Questions: 1. Decomposition Factor: (1-3b) a

2. Calculate 4 (1 2)- 8=0

Proof of: ab=de, a= d, af=dc triangle abf triangle dec

bf=ce,,∠afb=∠dce,∠cfb=∠fce, fb∥ce

The quadrilateral BCEF is a parallelogram.

2) i: BCEF is rhomboidal when af=7 5.

Solution: ac=5 from the Pythagorean theorem

Make BG ac at point G

then BG=12 5

cg=9/5

ag=16/5

So af=ag-cg=7 5

a minus 6ab plus 9ab squared = a (1-6b + 9b ) = a (1-3b) 3 calculation 4 half minus root number 8 = 2 2-2 2 = 0

Because af = cd, ab=de, and angle a = angle b, then the triangle abf is all equal to the triangle dec so ec=df, the angle dce=angle afb so the angle ecf = angle bfc, i.e. ec parallel bf so , the quadrilateral bcef is a parallelogram.

The angle abc is equal to 90 degrees, ab is equal to 4, and bc is equal to 3 connected to be intersected cf to g, when bcef is diamond, there is be perpendicular ef, so triangle abc is similar to triangle bgc

So ge = 3 5 be=

So af = ac- 2ge = 5

When af is , BCEF is diamond-shaped.

The triangle ABF congruence dec, bf = CE angle bfc = angle fce bf parallel and equal to CE, it can be proved to be a parallelogram.

af=2 bcef is diamond-shaped bf=3, fc=3, so af=2

Cheating in the high school entrance examination, this is not right, be careful across provinces, please go to tea, take the exam well, opportunistic shortcuts will eventually shoot yourself in the foot.

It's hard for him to type so many words.

There are still a few days left for the high school entrance examination.

First of all, is the shortest distance the shortest distance to walk from the inside of the cone or the surface of the cone?

According to the solution inside the cone: connect AB, make PD perpendicular to AB, and D is the center of the ground circle. If CE is parallel to PD, E is the midpoint of DB and CE is perpendicular to AB.

Solve the length of the hypotenuse ac in the right triangle ace, ae=ad+de=,ce2=(,ac=root(ae2+ce2)=? , do the math yourself.

The cone is fanned along the AP with a radius of 9 cm and an arc length of 2 factions R = 6 factions.

So the central angle of the circle is equal to (6 9 2) * 180 = 60 degrees.

At this point you get a triangle apc ap=9 pc= angular apc = 60 degrees.

It's easy to ask for ac below.,Do the math yourself.,The answer should be Root number 3 I'm a high school attached to the teacher.,If you don't understand, ask again.。

Shape the cone into a fan shape.

The length of the fan arc is the circumference of the bottom surface of the cone, which is 2 r=2* *3=6, and the fan angle is set as

6π=2πrθ/360=

Do AH vertical BP extension wire at H

apb=120°, then aph=60°

So ph=,ah=times root number 3

Pythagorean theorem, yield: ac = 9/2 times the root number 7

This is the question of the third year of junior high schoolFind the degree of the angle apb first. Then solve.

Solution: From the meaning of the question: a and b are the two roots of the equation x+1=x2, and the relationship between the roots and the coefficient can be obtained x1+x2=k, x1*x2=-1

That is, y1*y2=(kx1+1)(kx2+1)=k2 x1x2+k(x1+x2)+1

k2+k2+1

1 is constant, independent of k value (1).

That is, y1*y2=- x1*x2 is constant.

The slope score of OA is -1Equivalent to (y1 x1)*(y2 x2)=-1

That is, y1*y2=- x1*x2 is required to be consistent with the result of (1), so when k changes, there is always oa and ob

The bounty is not powerful!

Let's put it simply. Bring the coordinates of a b into the parabolic equation.

Combined with the slope formula, this can be demonstrated.

The first problem is a simultaneous equation. Just as the two equations are equal. List a quadratic function about x.

Use Vedant's theorem to find x1+x2 and bring in the equation to find it. Because 0 is the origin. It means that OA must be right to bring it in.

The second question should be discussed separately. It's too much trouble, you can do the math yourself. It's probably the same as the two-point distance formula.

I don't know.

There is difficulty, the first few are so difficult.

I'll answer, but the landlord said that ab, cd is an isosceles triangle waist??? I don't have a side.

22.（1）﹣1（2）0

Two-half root number two-by-half root number three one-half times two-half root number two = root number six root number two 4

sin15°=sin(45 30)=sin45cos30 cos45sin30=root number six root number two 4

sin 8cos 8=1 2 2sin 4cos 4=1 2cos ( 8) sin ( 8)=cos 4=2 of the root number of the second part.