I can t do junior high school math problems, and I often can t do junior high school math problems,

1. Greater than. 2、x+y=1

Process: Absolute value of x+3 + absolute value of y-4 = 0

So x+3=0 y-4=0

x=-3 y=4

So x+y=1

3、x=0,±1，±2，±3

Process: |x|<π

then the integer x=0, 1, 2, 3

4. The absolute value of x + the absolute value of y = 3

Idea: Two absolute values are opposite to each other, indicating that both sides are 0.

i.e. x=-1, y=2

So the absolute value of x + the absolute value of y = 3

5. The minimum absolute value of the equation x+1 is 0, in this case, the value of x is -16, and the answer is -3 2

Process: Because (x-1) 0, y+1 2 0 is constant.

And the square of the absolute value of x-1 + y + half = 0x-1 = 0, y + 1 2 = 0

x=1，y=-1/2

y-x=-1/2-1=-3/2

7. Solution: when x 5, 3-x 0, x-1 0, |3-x|=x-3,|x-1|=x-1

3-x|+|x-1|=x-3+x-1=2x-48, a=-8 b=2, or a=-8 b=-2 process: absolute value of (a-b) = b-a

Therefore, the absolute value of a-b<0 a and a = 8, and the absolute value of b = 2

Therefore a=-8 b=2 or a=-8 b=-2

9. Remove the absolute value symbol.

Original = [(1 99)-(1 101)]-1 99)-(1 100)]-1 100)-(1 101)].

Greater than 3-3, -2, -1, 0, 1, 2, 3 -1 is -8, b is -2

9.minus one hundred thousand two thousand one hundred and two.

That's pretty much it, pick me!

First read more examples, and then combine them to learn, and you will find that it is not that difficult!!

Summary. Hello dear, glad to answer for you, A.

If you want to do math problems well, you can only do them yourself, and you will never learn well by listening to others. After doing the questions, summarize more and build your own mind map. Focus on the foundation, grasp the question type, find the problem, and build bridges.

Return to the textbook, preview first, have a goal in listening, complete the homework carefully and refine the basis for solving the problem. First easy and then difficult, we must gradually achieve that no point is lost in simple questions, there are ideas for medium questions, and there are skills in solving problems. If you persist in this way for a period of time, your self-confidence will be there, and your fear of doing math problems will change, your mood will be better, and your grades will naturally improve.

There are often questions in junior high school math that I can't do, what should I do?

Dear, hello, I'm glad to answer for you, answer If you want to do math problems well, you can only do them yourself, and you will never learn well when you listen to others. After doing the questions, summarize more and build your own mind map. Focus on the foundation, grasp the question type, find the problem, and build bridges.

Return to the textbook, preview first, have a goal in listening, complete the homework carefully and refine the basis for solving the problem. First easy and then difficult, we must gradually achieve that no point is lost in simple questions, there are ideas for medium questions, and there are skills in solving problems. If you persist in this way for a period of time, your self-confidence will be there, and your fear of doing math problems will change, your mood will be better, and your grades will naturally improve.

Dear, if my answer is helpful to you, please give a thumbs up (comment in the lower left corner), look forward to your like, your efforts are very important to me, and your support is also the driving force for my progress. If you feel that my answer is still satisfactory, you can click on my avatar for one-on-one consultation. Finally, I wish you good health and a good mood!

Summary. Hello dear! Based on the question you described, Ms. Xiaoxi's answer is as follows:

Question 7: First find x1, then x -8x+4 -4 +10=0, then (x-4) =6, then x= 6+4 or x= 6-4. then substitute to:

x1+x2）/（x1x2）=（6+4+√6-4）/（6+4）*（6-4）=2√6/-10=-√6/5.

Hello dear, please send me the question, I will answer it for you.

Okay dear, wait a minute.

Dear, which question are you asking, question 8?

Okay dear, you wait.

Hello dear! Based on the question you described, Ms. Xiaoxi's answer is as follows: Question 7:

First, find x1, then x -8x+4 -4 +10=0, then (x-4) =6, then x= 6+4 or x= 6-4. Then substitution gets: (x1+x2) (x1x2)=(6+4+ 6-4) (6+4)*(6-4)=2 6 -10=- 6 5

Hello dear! According to the question you described, Mr. Xiaoxi's answer is as follows: 8. (a-1) +a-1) = (a-1) (a-1) = a(a-1) = a -a.

Hello dear! According to the question you described, Ms. Xiaoxi's answer is as follows: 13, 4 (m+2)=*m+2) 4=1 4+(m+2) (8-4m).

Summary. What's the question? Please send it over and see...

I can't do math problems, and it's more difficult in junior high school.

What's the question? Please send it over and see...

I'm sorry, I really can't do this question, I can't do it in a short time, in order not to delay your time. I'm really sorry.

-1 to 5 at a distance of 6

2 to -6 is 4 at a distance

1 to 12 is 11 at a distance

The distance from a to b is the absolute value of (a-b).

The distance between two points is the absolute value of (a-b).

First of all, to add the question, o is the center of the circle (the following solution ideas all require the condition that o is the center of the circle).

Connect OP first, and then pass P to make the perpendicular lines of AO and CO respectively, and intersect AO, CO and N respectively

At this time, you will find that the quadrilateral OCPA is divided into 2 triangles, AOP and COP, S triangle AOP=AO*PM*1 2=5*PM*1 2

S triangle cop=co*pn*1 2=5*pn*1 2

S quadrilateral ocpa=5*1 2*(pm+pn).

For ease of calculation, let pm=x,pn=y

Because ab is perpendicular to cd, pm perpendicular ab, pn perpendicular to pn, so quadrilateral pmon is rectangular, so mo=pn=y

Because po=5, then x 2+y 2=25

Because ao=5 and mo=y, am=(5-y).

Since ap=4, we can get x 2+(5-y) 2=16 by apm at right triangles

Combined with the above equation x 2 + y 2 = 25, we can calculate y=

It can also be deduced that x = 4/5 of the root number 21 = (4 5) * (21 (1 2)) Here is just a matter of expression, 1 2 times of 21 is the root number 21).

According to the S quadrilateral, OCPA=5*1 2*(PM+PN)=5*1 2*((4 5)*(21 (1 2))+17 5)=2*(21 (1 2))+17 2

First, there are two things to know:

1. The determination theorem of similar triangles: the two angles correspond to the equal, and the two triangles are similar.

2. The nature of similar triangles: the corresponding angles of similar triangles are equal, and the corresponding sides are proportional.

Schematic diagram of the solution.

Make a vertical auxiliary line to OA through the P point, and the two right-angled triangles APE and APB share A;

Using the determination theorem of triangle similarity, the ape apb can be derived;

From this, it can be deduced that ap ae=ab ap=pb pe is radiated by 5, ab=10;

From ap=4, ae= and pe= can be obtained

From the rectangular OEPF, OE=PF= can be obtained

So the area of the quadrilateral OAPC = the area of the OAP + the area of the OPC =

To make an auxiliary line, PE AB is crossed at the point E, the values of AE, OE and PE are obtained by using the triangle similarity, and the quadrilateral OCPA is split into a trapezoidal PEOC and a right-angled triangle APE, and then the area is calculated, as shown in the following figure.

Duyou, this can be made.

Because it is troublesome to draw, it can only be described in words.

Step 1: Connect the OP connection points, then the area of the quadrilateral OCPA is equal to the sum of the area of the triangle AOP and the triangle COP.

Step 2: In the triangle AOP, AO=PO=R=5, AP=4, you can find the sine value and cosine value of the angle AOP with trigonometric functions, you can find the area of the triangle AOP, AO*OP*SIN angle AOP* 1 2, you can find the area of the triangle COP, so you can find the area of the quadrilateral!

Double root number 21 plus.

Is that okay? Give a thumbs up if you agree. Please correct me if you don't make a decision, thank you.

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One of the respondents had the wrong answer, but it is the same as below.

If it's too cumbersome to enter, use **:

Solution: In RT APB as PE 丄ab

rt ape rt abp, ae 2 = ae be ae ap = ap ab = 4 10 i.e. ae = 8 5 pe = 4

eo=ao-ae=

Pass the P point as a PF CD and connect to the OP

pf=oe=

s quadrilateral oapc = s aop ten s opc

Please see ** for <> detailed steps.

<> the irregular graph to find the area, using the cut and patch method.

<> the division method to calculate, remember to give an oh, thank you!

Connect the OP, find the area of the triangle AOP and the triangle COP, and then sum it.

It can be made by P as PE perpendicular AB to E, PF perpendicular CD to F.

The lengths of PE and AE can be obtained by using AEP's similarity to APB.

The area of the triangle AOP is equal to one-half radius multiplied by PE, and PF = radius minus AE, and the area of the triangle COP is equal to one-half radius multiplied by PF, so that the final result is obtained.

It's so difficult, are all the math problems in junior high school so difficult now? I took a look at it and found that I wouldn't do it.

<> because AE is more difficult to calculate, it is not counted.

According to the solution, the vertical auxiliary line is made to OA through the P point, and the two right-angled triangles APE and APB share A;

Using the determination theorem of triangle similarity, the ape apb can be derived;

From this, it can be deduced that ap ae=ab ap=pb pe is radiated by 5, ab=10;

From ap=4, ae= and pe= can be obtained

From the rectangular OEPF, OE=PF= can be obtained

So the area of the quadrilateral OAPC = the area of the OAP + the area of the OPC =

The ratio of side lengths is 4:5, and the ratio of perimeters is also 4:5, so the circumference of two squares is 36 (4+5) 4=16, 36 (4+5) 5=20, and the side lengths are 16 4=4, 20 4=5, and the sum of the areas is 4 4+5 5=41

It is known that ab=10, ap=4, abp is a right triangle, pb= 84=2 21

ABP area = AP PB2 = 4 21

Since ABP obq, the corresponding edges are proportional: ob PB=OQ AP, 5 2 21=OQ 4

oq=10√21/21

obq area = ob oq 2 = 5 10 21 21 2 = 25 21 21

Area of the quadrilateral OCPA = ABP area - OBQ area = 4 21-25 21 21 = 59 21 21

Even OP as an auxiliary line, use the cosine formula to find the value of AOP, and then you can find the cosine value of the lost cop, and then use the cosine formula to find the length of PC.

Solution: S quadrilateral ocpa=s triangle AOP+s triangle POC=1 2OAXPM+1 2xOCXPN

1/2x5xpm+1/2x5xpn

5/2pm+5/2pn

5/2(pm+pn)

In the triangle AOP: OA=OP=5, AP=4COSAOP=(5 2+5 2-4 2) 2x5x5=17 25SINAOP=4x21 1 2 25

In the triangle opm, sinaop=pm op, pm=opsinaop=5x4x21 1 2 25=4x21 1 2 5

cosaop=om/op,om=opcosaop=5x17/25=17/5

s quadrilateral ocpa = 5 2x (4x21 1 2 5 + 17 5) = 2x21 1 2 + 17 2.

Answer: The area of the quadrilateral OCPA is 2x21 1 2+17 2.