2. In a square ABCD, m is the midpoint of AB, MN is perpendicular to MC, and BN divides CBE equally. Proof: MD=MN.
3. In triangle ABC, angle A = 100, and the bisector of angle B intersects with point E on the AC side. Verify BC=AE+BE.
4.ABCD is a square, P is the center of the square, and one side AD is the hypotenuse. Make a triangular AED outward and connect it with de, which proves that the DEA problem of PE bisecting angle is supplemented:
1. In the triangle ABC, the angle ABC = 2° angle C, and the bisector of the angle CBA intersects with AC at point D. Verification: AB+BD=AC]
The second question e is on the extension line to the right of AB.
The third question of triangle is not called isosceles triangle.
The leaning tower of the competition has eight floors and consists of more than 200 stone pillars (no more than 250). There are 12 at the top and 6 in the middle, with the same number of stone pillars on each floor. The number of stone pillars at the bottom is only half that of each middle floor, and the number of stone pillars at each floor and bottom is a multiple of 5. Find out the exact number of stone pillars on the leaning tower.
An old couple said, "The square difference of our ages is 195." A young couple said, "What a coincidence, the square difference of our ages is also 195." A middle-aged couple also leaned in and said, "The square difference of our ages is also 195." Please think about it. The three couples are of different ages.
"Grandpa often asks some strange questions," Ningning said. "In summer, my grandfather and I enjoyed the cool air on the roof of 10 meter high. Grandpa took the glass in his hand and said, I threw it into the sky, but it fell from 10 meters without breaking. Is this possible? "
"Is there anything soft downstairs, such as thick cotton or sponge?"
"No, it's an ordinary cement floor."
"Yes, this cup is special."
"No, it's an ordinary cup."
Do you know what this is？
Call for help ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The best answer is 1. If the number of stone pillars at the bottom is x, and each of the six middle floors is 2x, the total number of * *12x can be 12x+x+ 12.
2. Young couple 22, 17, middle-aged couple 3 1 34, old couple 98, 97. The solution can be listed as X 2-Y 2 = 195, 195 = 5 * 39 = 3 * 65 = 6545.
When x-y = 5,3, 1, x+y = 39,65, 195.
solve an equation
It is possible that it should be on the top of a ten-meter-high building. After being thrown into the sky, the height must be above 10 meter, so the cup didn't land when it fell to 10 meter, and it was still falling.
1. It is known that A and B are rational numbers and satisfy the equation 5-√2*a=2b+2/3√2-a,
Find the values of a and b
Supplement to the question:
2. Calculate the value of √11…11-222 … 22.
2n 1 n 2
3. Given that √a*a+2005 is an integer, find the sum of all positive integers A that meet the conditions.
4. Let the graphic area enclosed by two coordinate axes kx+(k+ 1)y= 1(k is a positive integer) be Sk(k= 1, 2,3, …, 2005), then S1+S2+S3+.
Answer (look carefully)
1.
5-√2*a=2b+2/3√2-a，
(2b-a-5)+(a+2/3)√2=0
Because a and b are rational numbers.
So 2b-a-5=0 and a+2/3=0.
The solution is a=-2/3 and b= 13/6.
2. Solution:
√( 1 1-2)=√9=3
√( 1 1 1 1-22)=√ 1089=33
√( 1 1 1 1 1 1-222)=√ 1 10889=333
………………………………
So √ (11…11-222 … 22) = 333 … 33.
There are 2n 1, n 2, n 2.
3. Solution:
Set √(a? +2005)=n, (n > a, and n, a is a positive integer)
And then a? +2005=n？
Is that n? -a？ =2005
(n-a)(n+a)= 2005 = 1 * 2005 = 5 * 40 1
So positive integers n-a= 1, n+a=2005, or n-a=5, n+a=40 1.
The solution is n= 1003, a= 1002, or n=203, a= 198.
So the sum of all positive integers A that meet the conditions is1002+198 =1200.
4. Solution: The intersection of straight line kx+(k+ 1)y= 1(k is a positive integer) and two coordinate axes is:
( 1/k，0)[0， 1/(k+ 1)]
The area of a closed figure.
sk = 1/2 * 1/k * 1/(k+ 1)= 1/[2k(k+ 1)]
So S 1+S2+S3+…+S2005.
= 1/2[ 1/( 1*2)+ 1/(2*3)+ 1/(3*4)+ 1/(4*5)+……+ 1/(2005*2006)]
= 1/2( 1/ 1- 1/2+ 1/2- 1/3+ 1/3- 1/4+ 1/4- 1/5+……+ 1/2005- 1/2006)
= 1/2( 1- 1/2006)
= 1/2*2005/2006
=2005/40 12
If a+ radical number 2ab+b= radical number 2 (ab is not in the radical number in the second term) and B is a rational number, then ()
A and A are integers B, A is rational number C, A is irrational number D, A can be rational number or irrational number.
The answer is C, but I don't know why it is C. Please answer!
The best answer is divided by ab.
1/b+√2+ 1/a=√2/ab
(a+b)/ab=√2(ab- 1)/ab
a+b=√2(ab- 1)
There are irrational numbers on the edge of the equation. If both A and B are rational numbers, the equation will never hold, and B is rational, then A must be irrational.