(x^2-7x)^2+6x-42x 25x^2-16y^2-4z+16yz

(x^2-7x)^2+6x-42x 25x^2-16y^2-4z+16yz

(x^2-7x)^2+6x-42
=[x(x-7)]^2+6[x-7]
=[x-7][x^2(x-7)+6 ]
=[x-7][x^3-7x^2+6]
=[x-7][x^2(x-1)-6(x^2-1)]
=[x-7][x^2(x-1)-6(x+1)(x-1)]
=[x-7][(x-1)(x^2-6x-6)]
25x^2-16y^2-4z^2+16yz
=25x^2-4[4y^2-4yz+z^2]
=25x^2-4[2y-z]^2
=[5x+2(2y-z)][5x-2(2y-z)]
=[5x+4y-2z][5x-4y+2z]

A square, the perimeter is 811 meters, the side length is___ M, the area is___ Square meters

(1) (2) 211 × 211 = 4121 (square meters) a: the area is 4121 square meters. So the answer is: 2114121

How to calculate the distance between the sun and the earth
There is no such long ruler, and the plane can't fly
People say it is measured by the speed of light. I am even more puzzled. How do people on earth know when light starts from the sun?

There are several methods to measure the distance between the sun and the earth. One is to use the transit of Venus (that is, the sun, Venus and the earth are just in a straight line); the other is to use asteroids to measure the distance between the sun and the earth. In history, the former method was used to measure the distance between the earth and the sun. Now the average distance between the sun and the earth is also calculated in this way, that is, a beam of radar wave is sent from the earth to hit Venus, The average distance between the sun and the earth measured by this method is 149597870 km, about 150 million km
In 1726, Halley proposed to measure the distance between the sun and the earth by observing the transit of Venus in different places
In 1677, 21-year-old Halley made a forecast of the transit of Venus in 1761. He knew that he could not see the transit of Venus in person. But Halley believed that as long as the apparent diameter of Venus was obtained by observing the transit of Venus, and the cycle of Venus was known, the solar parallax could be easily calculated by Kepler's third law, so as to calculate the distance between the sun and the earth
In 1761, as Halley had expected, there was a transit of Venus, but because the path of Venus was too close to the edge of the sun to measure accurately, astronomers had to complete Halley's ambition about eight years later at another transit of Venus in 1769
On May 23, 1769, under the cooperation of European astronomers and captain cook who sailed to Tahiti Island, accurate observation data were finally obtained. It is worth mentioning that Britain and France were at war at that time, but in order to complete this historic scientific exploration mission, the French government specifically ordered the Navy not only not to attack Captain Cook's endeavour, but also to prevent the Navy from attacking Captain Cook's endeavour, It is under this kind of international cooperation that the "astronomical units" which have not been solved for hundreds of years can be seen in the world under this rare opportunity of astronomical phenomena
In 1771, the French astronomer Lalande calculated for the first time the distance between the earth and the sun, which is about 152-154 million kilometers, based on this precious observation data, which is very close to the current measurement value of 149597870691 million kilometers
The modern distance between the sun and the earth is obtained by measuring the distance between the earth and Venus with radar
In the early 18th century, Halley proposed an ingenious method. I learned this method from Dazhong astronomy. It's hard to understand without a picture, but I can't find a picture, so I can only talk about it in general
First, the ratio of the distance between each planet and the sun to the distance between the sun and the earth can be determined by simple measurement
So just determine the distance between a planet and the sun. The question is how
Harley thought, Venus transit, sun, gold, earth three body line, but from different places, Venus in the sun's projection is not the same! This is not difficult to explain, you can think, the moon in the total solar eclipse on the ground projection is also different, only in a small point there will be a total solar eclipse. Similarly, Venus is the same, but the difference is much smaller than the moon
During the transit of Venus, we observe from the southern hemisphere and the northern hemisphere respectively, record the projection of Venus on the solar surface, and then calculate to get the ratio of the distance between Venus and the earth and the straight-line distance between the two observation points. (in fact, measuring the transit time of Venus is more useful, and the error is smaller. I can't explain it without a picture. I hope you can think about it yourself.)
It is relatively easy to measure the radius of the earth (the specific method is omitted), so it is not difficult to get the straight-line distance between two observation points by simple calculation. With this data, the earth gold distance can be calculated, and then the Earth Day distance can be obtained
The method of measuring the distance between the earth and the moon is roughly the same, but the background is replaced by the stellar background, so the measurement is relatively easy. Basically, the distance between the earth and the moon is obtained as soon as the radius of the earth is obtained

When a train departs from the station and makes a uniform acceleration linear motion with zero initial speed, and the displacement is 0.5m in the third second, then the displacement is () m in the middle and back of 1min, and the instantaneous speed is () m / s
The answer is 360m 12m / s

I'm a high school physics teacher. It's a little difficult to understand. The displacement in the third second is 0.5m, so the average speed in the third second is v = s / T = 0.5 / 1 = 0.5m/s, the speed in the middle is the average speed, so the speed in the second second second is 0.5m/s. According to v = at, so 0.5m/s = a x 2.5, so a = 0.2m/s & sup2

Given that the area of a square is (16-8x + x2) cm2 (x > 4cm), the perimeter of the square is ()
A. （4-x）cmB. （x-4）cmC. （16-4x）cmD. （4x-16）cm

∵ 16-8x + x2 = (4-x) 2, x > 4cm, the side length of the square is (x-4) cm, the perimeter of the square is: 4 (x-4) = 4x-16 (CM), so D

Simple calculation (90 + 1 / 88) × 1 / 89
sit back and wait!

(90 + 1 / 88) × 1 / 89
=(89 + 89 / 88) × 1 / 89
=89 × 1 / 89 + 89 × 1 / 88
=1 + 1 / 88
=1 and 1 / 88

Find the following function range y = the square of (lgx) under the root - the square of lgx (100 < x less than 1000) y = with 7 / 2
Find the range of the following functions
Y = the square of (lgx) under the root - the square of lgx (100 < x < 1000)
Y = based on 7 / 2 (6-5x-x) (- 3 ≤ x < 0)

1. Let t = lgx, then 21
So the range of Y is ((7 / 2) ^ 6, (7 / 2) ^ (49 / 4)]

The edge length of a square is four centimeters. If you dig out a square with an edge length of two centimeters, it is necessary to calculate the surface area and volume of a large square

As the first floor said: volume = 7 / 8 of the original volume = 56 cubic centimeters
Surface area, if you dig from the corner, the surface area is the same as the original = 96 square centimeter; if you dig from the side, you need to add two small cube faces = 104 square centimeter; if you dig from the face, you need to add four small cube faces = 112 square centimeter; if you dig from the center (it's not easy to dig, it's the sum of two cube faces = 120 square centimeter)

5. 8, 6 and 2 are equal to 24 by calculation

The first algorithm: (8-5) * (2 + 6) = 24
The second algorithm: 5 * 6-8 + 2 = 24

It is known that all items of the sequence {a} are positive numbers, and N and Sn satisfy 6sn = an square + 3an + 3. If A2, A4 and A9 are equal ratio sequence, the general term formula can be obtained

6Sn=an^2+3an+3,6S(n-1) = [a(n-1)]^2 + 3a(n-1) +3
Phase difference: 6An = an ^ 2 + 3an - [a (n-1)] ^ 2 + 3A (n-1) (Note: Sn - S (n-1) = an)
The results show that: (an-a (n-1) - 3) * (an + a (n-1)) = 0
So: an - A (n - 1) = 3
The sequence is an arithmetic sequence with a tolerance of 3
6sn = an ^ 2 + 3an + 3, when n = 1, 6A1 = A1 ^ 2 + 3an + 3
I did it for nothing. This problem is wrong. I can't find A1