(one square of 1-2) times (one square of 1-3)... Times (one square of 1-9) times (one square of 1-10) equals?

(one square of 1-2) times (one square of 1-3)... Times (one square of 1-9) times (one square of 1-10) equals?

Let an = 1-1 / (n + 1) &# 178; = (n & # 178; + 2n) / (n + 1) &# 178; = n * (n + 2) / (n + 1) &# 178; so an + 1 = (n + 1) * (n + 3) / (n + 2) &# 178; so an * an + 1 = n * (n + 3) / * (n + 1) (n + 2) (1-2's Square) times (1-3's Square)... Times (1-9's Square) times (1-10's Square)

Square of X + 12x = 27

x^2+12x-27=0
It can't be factorized

The square of X + 12x + 25 = 0 is solved by factorization

x²+12x+25=0
△=12²-4×25=44
∴x=(-12±2√11)/2=-6±√11
It can't be factorized
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When the vehicle runs at a speed of 10m / s and gets an acceleration of 2m / s after braking, the distance of passing 4 s after braking is? 8 s, and the distance of passing 8 s is? 8 s?

There is a trap in this problem. Note t = V / a = 5S, that is to say, the car stops 5 seconds after braking
According to the displacement formula s = v0t + 1 / 2at ^ 2
1、S1=10*4-1/2*2*4^2=24m
2. The car stops at the 5th second: S2 = 10 * 5 -- 1 / 2 * 2 * 5 ^ 2 = 25m
Understand?

Xiaoming and Xiaogang run the 100 meter race. Xiaoming runs faster than Xiaogang. If they start at the same time, Xiaoming will win. Now Xiaoming lets Xiaogang run 10 meters first and catch up with Xiaogang at the start. As shown in the figure, L1 and L2 represent the relationship between the distance s (m) they run and Xiaoming's catch-up time t (s) respectively. Who will win the race?

If Xiaoming's speed is V1 and Xiaogang's speed is V2, it will take t = 100 / V1 for Xiaoming to run to the finish line, and at this time, Xiaogang takes v2t + 10 = 100v2 / V1 + 10 to calculate the difference. If Xiaogang runs more than 100 meters, Xiaogang will win. If it is less than 100 meters, Xiaoming will win 100 - (1

Given that one root of the equation X2 - [k2-9] + k2-5k + 6 = 0 is less than 1 and the other root is greater than 2, find the value range of the real number k?

X^2-(K^2-9)+K^2-5K+6=0
X ^ 2 = 5k-15 > = 0
Then: x = - radical (5k-15) (2)
From (1): 219 / 5

On the number axis, points a and B represent rational numbers a and B respectively, and the origin o is just the midpoint of AB, then the value of 26 / 1995a × 3b is to seek the help of the great God

On the number axis, points a and B represent rational numbers a and B respectively, and the origin o is just the midpoint of AB, which indicates that a and B are opposite numbers, so 1995a multiplied by 26 / 3B = - 17290

Using proportion to solve mathematical problems
There are two piles of medium a and B, and their mass ratio is 6:5. If 560kg is transported from pile a to pile B, the mass ratio of two piles of medium a and B becomes 2:3. How many kg is the original pile a? Please tell me the calculation process!

If the original mass of pile a is 6 x before transportation, the mass of pile B is 5 X
The mass of pile a after transportation is 6x-560, and that of pile B is 5x + 560
So (6x-560) / (5x + 560) = 2 / 3
3（6x-560)=2(5x+560)
18x-1680=10x+1120
8x=2800
X = 350 (kg)
Therefore, before transportation, the original mass of pile a was 2100Kg, and that of pile B was 1750kg

Who has the Mathematical Olympiad application problem of the equation of one degree, more difficult
In the difficult point

A cyclist walks at a speed of 18 kilometers per hour at first. He starts at a place where the rest of the journey is 32 kilometers less than the one he has already traveled. He walks the whole journey at a speed of 25 kilometers per hour. If the average speed of the whole journey is 20 kilometers per hour, how many kilometers did he walk?

Square B minus 4b of factorization a

a^2b-4b=b(a^2-4)=b(a+2)(a-2)