Calculation: (1) (- 2XY & sup2;) * (3x & sup2; y) & sup2; (1).(-2xy²)*(3x²y)²;(2).(-a)²(2-3ab);(3).(x+2y)(2x-y);(4).(a-2b)(a+2b);(5).(2a+1)²;(6).(a-3b)² Come on, wait online for 30 minutes

Calculation: (1) (- 2XY & sup2;) * (3x & sup2; y) & sup2; (1).(-2xy²)*(3x²y)²;(2).(-a)²(2-3ab);(3).(x+2y)(2x-y);(4).(a-2b)(a+2b);(5).(2a+1)²;(6).(a-3b)² Come on, wait online for 30 minutes

(1).(-2xy²)*(3x²y)²
=(-2xy²)*(9x^4y²)
=-18x^5y^4
(2).(-a)²(2-3ab)
=a²(2-3ab)
=2a²-3a³b
(3).(x+2y)(2x-y)
=x(2x-y)+2y(2x-y)
=2x²-xy+4xy-2y²
=2x²+3xy-2y²
(4).(a-2b)(a+2b)
=a²-(2b)²
=a²-4b²
(5).(2a+1)²
=(2a)²+2×2a×1+1²
=4a²+4a+1
(6).(a-3b)²
=a²-2a×3b+(3b)²
=a²-6ab+9b²

Calculation: 3x & sup2; y × (- 2XY & sup3;)

3x²y×（-2xy³）=-6x^3y^4

The fourth power of (2) (- 2XY & sup2;) & sup3; = (); (- 2x & sup2; y) * 3x is calculated

: (2) (- 2XY & sup2;) & sup3; = (- 2x ^ 3Y ^ 6); (- 2x & sup2; y) * 4x power of 3x = - 6x ^ 6y

How to use Taylor formula to find limit?
For example, how can LIM (x – > 0) {1 + 1 / 2 (x ^ 2) - (1 + x ^ 2) ^ (1 / 2)} / {(cosx-e ^ (x ^ 2)) sin (x ^ 2)} be broken?

It is to remember the expansion of the five or six basic functions. When encountering similar function limits, if the equivalent infinitesimal and the law of Robida are not easy to use or more complex, we can consider Taylor series expansion to find the limit. As for the order of expansion, it generally depends on the numerator or denominator. If it is the expansion of two additive or subtractive functions, it is expansion, Until the infinitesimal whose coefficient is not zero appears
lim(x–>0){1+1/2(x^2)-(1+x^2)^(1/2)}/{(cosx-e^(x^2))sin(x^2)}
First of all, the (1 + x ^ 2) ^ (1 / 2) term in the molecule needs to be expanded. Because there is also the 1 + 1 / 2 (x ^ 2) term in the molecule, you only need to expand it to the fourth power term of X. This is what I said before until the term with non-zero coefficient appears
Then, because the numerator is equivalent to x ^ 4 / 8, the denominator is also close to this direction. Because there is a sin (x * x) in the denominator that is equivalent to x ^ 2, the preceding cosx-e ^ (x ^ 2) of course only needs to be expanded to the second power of X
Because cosx ------ 1-0.5x * x
e^x---------x
The answer should be - 1 / 12

The number of female students is four fifths of that of male students, and female students account for four ninths of the total number of students in the class（

5:4

A = B times 57 A times 2 equals B times () a times 5 equals B times () a times () equals B times 12 a times () equals B times 29
Please tell me!

A = B times 57
A times 2 equals B times (114);
A times 5 equals B times (285);
A times (12) equals B times (57 * 12);
A times (29) equals B times (57 * 29)
Do it yourself

Write a general formula of the following sequence: 1, 0, - 13, 0, 15, 0, - 17, 0

When n is even, an = 0; when n is odd, if n = 4K + 1, then an = (- 1) n + 1 · 1n; if n = 4K + 3, then an = (- 1) n · 1n (K ∈ n)

5 / 39 × 55 / 26 △ 3 / 11 (simple calculation) 4 / 5 / 3 - 1 / 3 × 1 / 7 / 2

5／39×26／55÷ 3／11
=5／39×26／55 ×11／3
=2／9
4 and 3 / 5 - 1 / 3 × 1 / 7
=23／5-1／3× 9 ／7
=23／5-3／7
=146／35

Find the rules of junior high school (1) 1,9,25,49;
(1)1,9,25,49, , ;（2）2/3,3/9,4/27, , ;(3)1,-2,4,-8, , .

（1）9-1=8
25-9 = 16, so it's the difference × 2
(2) Numerator + 1 denominator × 3
(3) Multiply each number by - 2

For any prime number P, we prove that there are infinitely many positive integers n such that P can be divided by (2 ^ N-N)

From Fermat's theorem, we can get P | 2 ^ (p-1) - 1, so p | 2 ^ (p-1) - 1-p = 2 ^ (p-1) - (P + 1) so let n = K (P ^ 2-1), then 2 ^ n = [2 ^ (P ^ 2-1)] ^ k = [2 ^ (p-1)] ^ (K (P + 1)) = (- 1) ^ (K (P + 1)) = 1 (mod p) 2 ^ n = 1 - K (P ^ 2-1) = 1 + K (Mo