2x^2-3x=12

2x^2-3x=12

2x^2-3x-12=0
Cross multiplication is not true, only the root formula can be used
x=(-b±√ b^2-4ac)/2a
a=2,b=-3,c=-12
x=(3±√105)/4

If 3x and 4-x are opposite to each other, what is x equal to?

3x=x-4
x=-2

Ancient poems describing the Three Gorges of the Yangtze River
Write down the source and author

The poem about the Three Gorges: 1. The song of emperor Bai in the dynasty, the ape on both sides of the Qianli River mausoleum can't stop singing in one day, and the light boat has passed the heavy mountain. 2. The nine peaks are seen on the 12th Wushan mountain, and the bow of the boat is covered with green clouds in the autumn sky. The ape cries in the moon all night. 3. The boat turns to the mountain, and the white waves cross the river before it arrives

Taylor formula limit, how to know is the expansion order?
lim x->0 [ (e^x)*(sinx)-x(1-x) ] / (x^3)
In the book, e ^ X and SiNx are expanded into third order, but they are multiplied. Why are they expanded into third order?
In addition, the detailed explanation of this problem, especially the expansion of the multiplication of SiNx and e ^ X

The latter part of the molecule is x-x ^ 2. Since it is only quadratic, the former e ^ x * SiNx only needs x ^ 3. Maybe the x ^ 2 term can't offset it. So expand e ^ X and SiNx to the third order and multiply them. E ^ x = 1 + X + 1 / 2 * x ^ 2 + 1 / 6 * x ^ 3 + O (x ^ 3) SiNx = X-1 / 6 * x ^ 3 + O (x ^ 3) e ^ x * SiNx = x + x ^ 2 +

Girls account for five ninth of the class, while boys account for four fifths of girls. Right or wrong?

If the number of men is x, the number of women is 5 / 9x, then the number of men is 4 / 9x. Are there more boys or more girls? Besides the number of girls, the number of boys is equal to the number of girls. Isn't it 4 / 5? If you understand, you will know that the answer is right?

69 × 99 + 67 × 2 = simple method

69×99+67×2
=69×100-69+67+67
=6900+65
=6965

How to get the general term formula and the sum of the first n terms of the sequence 1,2 + 3,4 + 5 + 6,7 + 8 + 9 + 10

In fact, it is easier to find Sn first and then an
SN is obviously added from 1 to a certain number
Find the law
N = 1, the last number is 1
N = 2, the last number is 1 + 2 = 3
N = 3, the last number is 1 + 2 + 3 = 6
So the last term of Sn's 1 + 2 +... + m is 1 + 2 +... + n = n (n + 1) / 2
So Sn = 1 +... + n (n + 1) / 2
=[1+n(n+1)/2][n(n+1)/2]/2
an=Sn-Sn-1
=[1+n(n+1)/2][n(n+1)/2]/2-[1+(n-1)(n)/2][(n-1)(n)/2]/2
=[n+n^3]/2

Calculation: 1 / 55 + 2 / 55 + 3 / 55 +. + 54 / 55

1/55+2/55+3/55+.+54/55=55*27/55=27

Find the rules 1, 4, 9, (), 25, (), 49

1、4、9、（16 ）、25、（36 ）、49
an=n²

It is proved that when n > 1, there is no odd prime P and positive integer m such that P ^ n + 1 = 2 ^ m; when n > 2, there is no odd prime P and positive integer

If n is even, let t = P ^ (n / 2), then T ^ 2 + 1 = 2 ^ M. because n > 2, P > = 3, so m > 3. T ^ 2 + 1 = 2 ^ m, mod4: T ^ 2 = 3 (mod4) contradiction. If n is odd, then 2 ^ m = P ^ n + 1 = (P + 1) (P ^ (n-1) -. + 1). So there is k > = 2, such that P = 2 ^ k-1. SO 2 ^ M-1 = (2 ^ k-1) ^ n (obviously m > k) = 2 ^ kn -. + n * (2 ^ k)