Junior high school mathematics competition questions, junior high school mathematics competition que

The 4th floor is correct, and I'm in detail.

Solution: 1. Original formula = (x 5+1 x 5) + (x 5+1 x 5) 2-2....It's complicated to break it down).

2. From the known (x+1 x) 2=3*3 .Square on both sides of the equation) x 2+1 x 2=7 ..

x+1/x）（x^2+1/x^2）=7*3x^3+(x+1/x)+1/x^3=21

x^3+3+1/x^3=21 ..Here's the point of substituting the known into ) x 3+1 x 3=18 ...

3. Formula * De.

x 2 + 1 x 2) (x 3 + 1 x 3) = 7 * 18 simplified to obtain x 5 + 1 x 5 = 123 .

4. Put the substitution in.

x 2+1 x 2+2=9 squared on both sides

i.e. x 2+1 x 2+2=7....1

Multiply by the known conditions to get x 3+1 x 3=18....Multiply 21 by 2 to get x 5 + 1 x 5 = 18 * 7-3 = 123....33 squared, x 10 + 1 x 10 + 2 = 123 * 123 x 10 + 1 x 10 = 15127

x^10+1/x^10+x^5+1/x^5=15250

x+x/1=3 (x/x, I denote a).

Original = (x+a) to the 5th power -2 + (x+a) to the 10th power - 2

3 to the power of 5 + 3 to the power of 10 - 4

Solution: Because x+1 x=3 i.e., x=3-1 x bring it into and solve it!

1. A total of 180.

There are 900 three-digit numbers, from 100 to 999, the 900 numbers are arranged from small to large, every ten is a group, divided into 90 groups, for any group, let the sum of the three digits of the first number be a, then the next one is a 1, a 2 ,..a 9, which is 10 consecutive natural numbers, and only 2 of the 10 consecutive natural numbers are divisible by 5, so there must be and only 2 of these 10 numbers meet the condition, so there are a total of 2 90 180 numbers that meet the condition.

For example: 310, 311, 312 ,..319, the sum of the numbers of this set of 10 numbers is 4, 5, 6 ,..13, of which only 5 and 10 are divisible by 5, i.e. 311 and 316

2. Let's take a look at the following rules first.

11 = 121, and the sum of its digits is 1 2 1 = 4

111 = 12321 and the sum of its digits is 1 2 3 2 1 = 9

1111 = 1234321, and the sum of its digits is 1 2 3 4 3 2 1 = 16

So 11....The sum of the digits squared of 11 (1989 1) is 1 2 3 4 ...1989＋1988＋..

Substituting ab in, c is equal to 1, the relationship between a and b is known, s is equal to 2b + 2, and then according to the first quadrant, to find.

Start by drawing a picture.

Highlight two dots.

By fixed point in the first quadrant.

That is, find a random spot in the first quadrant.

It is then connected with known points to form a parabolic chart.

That is, it can be known that the opening is downward a<0

c=1b>0

Bringing the point b into the original equation yields a-b+c=0

Because c=1, a-b+1=0

Move to b=a+1

So s=a+b+c=2a+2

Because a>0, 2a>0

2a+2>2

So s>2

Decompose 2001 into prime factors.

It can form a triangle, and the sum of the lengths of any two sides is greater than the length of the third side, and then slowly make up.

There are 7 of them:

The matchstick with the circumference of the nth major triangle is 3n

The number of small triangles is n 2

A total of n -3n is required

The prime factor for 2001 is only three numbers.

Because it does not satisfy either side and is greater than the third side, it cannot form a triangle.

Actually, looking at the small triangles, the first one is a triangle, the second one is 1 in a row, 2 in a row, the nth is 1 in a row, 2 in the second row, and n triangles in the nth row, so the number of triangles in the nth figure is (1+2+. n) pcs, while each triangle consists of three matchsticks.

Therefore, the number of matchsticks is 3 (1+2+..)n) is 3n*(n+1) 2.

Verify as follows. The first 3*1*(1+1) 2=3.

The second 3*2*(2+1) 2=9.

1.The hundredth digit can be taken 1 9, the ten digit can be 0 9, and the single digit can only be two after the high two digits are determined. 9*10*2=180

2.It is approximately equal to (1989+1988)*5=19885, and the exact number needs to be considered in detail.

Consider the carry problem.

Upstairs and downstairs think about it, 11....What are the digits of the square of 11 (1989 1s), which should be 1989*2-1=3977 digits. Even if every bit is 9, it's just 35793, so the answer won't be 1989 2=3956121

AD is the angular bisector of the triangle ABC, may I ask which angle bisector of the angle is he?

From the question, we know that after removing the absolute value sign, x should be eliminated.

It is observed that 2+3+4+5+6+7+8+9+10=54, and 8+9+10=27, so when 7x is less than 1 and 8x is greater than 1, it means that x is less than one-seventh but greater than one-eighth.

1-2x）+（1-3x）+（1-4x）+（1-5x）+（1-6x）+（1-7x）+（8x-1）+（9x-1）+（10x-1）=6-3=3

It is equivalent to adding a plus or minus sign before (1-kx), which is a constant to indicate that the coefficient of x is 0;

From the known coefficients, it is known that all negative signs are from a certain beginning, and all positive signs are before it.

i.e. (1-2x) + (1-3x) + ....1-7x）-（1-8x）-（1-9x）-（1-10x）=3

In order to satisfy that p is a constant value, that is, the absolute value x can be reduced to be removed.

2x-3x-4x-5x-6x-7x+8x+9x+10x=0, that is, 1-7x>=0 and 1-8x<=0

When 1|8<=x<=1|At 7, p is a constant value p=3

Unit 8: 1*8 2*4 6*8

So the probability is 8 single digits of 1.

2 single digits are 6.

4 single digits of 2.

6 single digits are 8.

8 single digits are 6.

Because ask n maximum.

So from the consideration that 8 single digits are 1 and 8 single digits are 6, if 8 single digits are 1:

If you erase all the single digits of these n numbers, then the sum of these numbers is (2008-8) 10=200

Substituting the formula for summing the difference series:

sn=na1+n(n-1)d/2

200=8a1+8*7*6 2 a1=4 then this situation is satisfied.

i.e. n maximum is 8

The question is that it can't be all odd numbers, right?

First, prove that the square of the odd number is divided by 8 and the remainder is 1.

For an odd number 2k+1 (k is the whole spring number), the potato stove is (2k+1) =4k(k+1)+1, and k,k+1 is two consecutive integers, so 4k(k+1) is divisible by 8, so the odd square is divided by 8 and the remainder is 1.

Because the remainder of an odd number divided by 8 is 1, if all 6 numbers are odd, then the remainder divided by 8 on the left is 5, and the remainder divided by 8 on the right is 1, and the two sides cannot be equal.

So these six numbers can't all be odd.