Olympiad topics in the second year of junior high school, quadratic radical type.

Original = (6 + 3) + 3 ( 3 + 2) ( 6 + 3 ) ( 3 + 2).

So the original = 4030055

3）1/〔（n+1)√n+n√（n+1）=√n/n-√（n+1）/（n+1）

Original = 1- 2 2+ 2 2- 3 3+ 3 3 3 -..99/99-√100/100

4) Original = ( 3 + 1) 2 + 3-1) 2

2. From the root number, we can know a>1, then 1-a<0, move into the root number, then the original formula =- [a-1) 2*1 (a-1)]=- (a-1).

1. Move one to the other side, square both sides, and get a+4=25-10 (a-1)+a-1

a-1)=2, get a=5

Then 6-2 a = 6-2 5 = 5-1

5. x+y=-6, xy=3, the judgment is x, y minus, x+y) 2=36=x 2+y +2xy

x^2+y^2=36-6=30

y/x)+√x/y)]^2=y/x+x/y+2=(x^2+y^2)/xy+2=30/3+2=12

6、x=13/√（4+√3）^2=13/(4+√3)=4-√3

That's it for now.

Hope it helps. It's too time-consuming.

If you can't see clearly, you zoom in a little bit, or give an email address or QQ

I'll do it if I have more bounty points.

It doesn't matter who does it for you!

Besides, there are so many topics.

Mathematics Olympiad for junior high school Introduction] Mathematics Olympiad or Mathematics Olympiad, referred to as Mathematics Olympiad. The Olympiad embodies the commonality between mathematics and the spirit of Olympiad sports: faster, higher, stronger.

As an international competition, the International Mathematical Olympiad is proposed by international mathematics education experts, and the scope of questions exceeds the compulsory education level of all countries, and the difficulty is much more than that of university entrance exams. Olympiad mathematics has a certain effect on the mental exercise of teenagers, and can exercise thinking and logic through Olympiad mathematics, which does not only play a role in mathematics for students, but is usually more esoteric than ordinary mathematics. The following is the knowledge points of the Olympiad in the second grade of junior high school for everyone:

The operation of quadratic radicals is mainly the study of multiplication, division, addition and subtraction of quadratic radicals.

1) Addition and subtraction of quadratic radicals:

It is necessary to simplify the quadratic radical first, and then add or subtract the coefficients of the quadratic radical with the same number of open squares (i.e., the same quadratic radical), and the number of opened squares remains unchanged.

Note: For the addition and subtraction of quadratic radicals, the key is to merge the same quadratic radicals, usually into the simplest quadratic radicals first, and then merge the same quadratic radicals before the group. However, when simplifying the quadratic radical, the number of the square of the quadratic radical should not contain a denominator and a factor that can be opened to the end.

2) Multiplication and division of the quadratic root argument search:

Note: The operation rules of multiplication and division should be used flexibly, and in actual operations, they are often transformed from the right side of the equation to the left side of the equation, and at the same time, the value range of the letters should be considered, and finally the operation result should be reduced to the simplest quadratic radical.

3) Quadratic radical hybrid operations:

Multiply (or open) first, then multiply and divide, and finally add and subtract, and those with parentheses are counted in parentheses first; If you can use the arithmetic law or multiplication formula to perform the operation, you can appropriately change the order of operation to perform simple operation.

Note: When performing radical operations, it is necessary to correctly use the algorithms and multiplication formulas, analyze the characteristics of the problem, and master the methods and skills to make the operation process simple. The results of quadratic radical operations should be kept as simple as possible.

In addition, fractions of the radical formula must be written as false fractions or true fractions, not as fractions with fractions.

2) +1 rutin (3 2 + 2 3) +....1/（100√99

The process is as follows: the first step: move the coefficient outside the root number of the denominator to the root number, and then extract the common factor such as.

For example, the denominator (3 2 + 2 3) = the square of 3 under the root number 2 + the square of the root number 2 3 = ( 3 2 ) ( 3 + 2).

Each item does this, and the last denominator :

99 100 = 100 squared under the root number 99 + 99 squared by the root number 100 = (100 99) (100 + 99).

Step 2: The addition part of the formula extracted in the previous step, the denominator is rationalized, such as 1 [(3 2)( 3 + 2)]=

Each of them does so.

So the original formula is equal to =++1-(1 2)]+1 2)-(1 3)]....1/√99)-（1/√100)]

The addition and subtraction of the middle terms are eliminated and only the first and last terms remain, so.

Original formula = 1 - (1 100) = 1 - (1 10) = 9 10

ab+a-b=1,ab+a=b+1

a(b+1)=a

Since a is an irrational number, b+1=0 and b=-1

a(b+1)-(b+1)=0

a-1)(b+1)=0

Because a is an irrational number, a-1 is not 0

So b=-1

1 x1+1 x2=(x1+x2) x1x2=-b c=-3 (-m+1)=3, so m=2. I wish you progress!

Collapse of the quarrel: Let the two poles of 2x +3x-m+1 = 0 respectively rise to judgment circle x1 and x2, then rush up:

x1+x2=-3/2; x1*x2=(-m+1)/2.

1 x1+1 x2=3, i.e., (x1+x2) (x1*x2)=3,(-3 2)[m+1) 2]=3,m=2

2x +3x-m+1=0 in a=2, b=3, c=1-m set two as x1, x2, then there is a crash.

1/x1+1/x2=3

and 1 x1+1 shirt x2=(x1+x2) (x1x2)=(b a) (c a)=-b c=-3 (1-m)=3 to get m=2

Let the two roots be x1 and x2 respectively

Then there is x1+x2=-3 2, x1·x2=(-m+1) 2 by the meaning of the title. 1/x1+1/x2=3

Then the wheel is divided into (x1+x2) (x1·x2)=3, that is, 3 (m-1)=3

Ulalade m = 2

Spend 20 or 30 dollars to find a bookstore and buy two good copies.

Find how many this equation is equal to, and a is greater than 0

Actually, think about it again. A question suddenly came to mind...

Passing C as the perpendicular line of AB, crossing AB at point E, crossing the perpendicular line of D left CE, crossing CE at point G, then the angle B=60 degrees, BC=2BE, the angle CDA=120 degrees, the angle CDF=30 degrees, so DC=2CF, so CF=1, DF=root number 3

So eb=2 root number 3, so bc=4 root number 3, so ce=6, so da=5 so the perimeter is 7+7 root number 3

Extend the intersection of AD and BC at E, A= C=90°, B=60°, so E=30°.

cd=2, ab=3, root number 3, so ec=2 root number 3, eb=6 root number 3, bc=4 root number 3, de=4, ae=9

So ad=5, ab+bc+cd+da=7, root number 3+7

From cd=2, df=3,cf=1 and ab=3 3, so be=2 3, bc=4 3, ce=4, so ad=ef=ce-cf=4-1=3

So the perimeter of the quadrilateral = ab + bc + cd + ad = 3 3 + 4 3 + 2 + 3 7 3 + 5

(1) Root number x 2+2ax+a 0 |x-a|0 and their sum equals 0, which means that they are all equal to 0, so x=a solves a 2+2a 2+a=0

a = 0 or -1 3 (2).