1993 + 1992 times 1994: 1993 times 1994 minus 1

1993 + 1992 times 1994: 1993 times 1994 minus 1

(1993*1994-1)/(1993+1992*1994)=(1993*1994-1)/[1993+(1993-1)*(1993+1)]=(1993*1994-1)/[1993+(1993^2-1)]=(1993*1994-1)/[1993+1993^2-1]=(1993*1994-1)/[1993*(1+1993)-1]=(1993*1994-1)/[1993*1994-1]=1

Given the sequence an, an = 1 / (n (n + 1) × (n + 2)), then LIM (n tends to be infinite) Sn

an=(1/2)[1/n(n+1)-1/(n+1)(n+2)]
So Sn = 1 / 2 [1 / 1 * 2-1 / 2 * 3 + 1 / 2 * 3-1 / 3 * 4 + +1/n(n+1)-1/(n+1)(n+2)]
=1/2[1/2-1/(n+1)(n+2)]
N tends to infinity and then to zero
So limit = 1 / 2 * 1 / 2 = 1 / 4

How to solve [3 / 7X + 1 / 4x + (3 / 7X + 1 / 4x-15)] - 115 = x

[3/7x+1/4x+(3/7x+1/4x-15)]-115=x
3/7x+1/4x+3/7x+1/4x-15-115=x
6/7x+1/2x-130=x
6/7x+1/2x-x=130
5/14x=130
x=130÷5/14
x=364

A barrel of oil weighs 12.65 kg. After half use, the barrel weighs 6.85 kg. How many kg?

12.65 - (12.65-6.85) × 2, = 12.65-5.8 × 2, = 12.65-11.6, = 1.05 (kg); answer: the barrel weight is 11.05 kg

If the piecewise function f (x) = e ^ x + 2x ^ 2-x + 1 (x is not equal to 0) f (x) = K (x = 0) is continuous within (negative infinity to positive infinity), then the value of K is

The function f (x) = e ^ x + 2x ^ 2-x + 1 is continuous,
So the limit at 0 is the function value limf (x) = 1 + 1 = 2, X tends to 0
And f (0) = K
∴k=2

2.5x-1.8 + 0.5x = 0.615.6-2 (x-7.4) = 12.40.86 * 2-7.1x = 0.335-x = 4 (x + 5)

2.5x-1.8 + 0.5x = 0.63x = 0.6 + 1.8x = 0.815.6-2 (x-7.4) = 12.415.6-2x + 2 * 7.4 = 12.4-2x = 12.4-15.6-2 * 7.4-2x = - 18x = 90.86 * 2-7.1x = 0.3-7.1x = 0.3-1.72 (product of 0.86 * 2) - 7.1x = - 1.42x = 0.235-x = 4 (x + 5) 35-x = 4x + 4 * 535-20 = 4x + x5x = 15x = 3

The volume of a pool is 90m & sup3;, and the current water storage is 10m & sup3;, and water is injected into the pool by water pipe at the speed of 5M3 / s

Fill up after X hours
Existing water volume + water injection per hour × water injection time = pool volume
10＋5×x＝90
10＋5x＝90
5x＝90－10
5x＝80
x＝16

Does the Equivalent Infinitesimal Substitution when x tends to 0 also apply when x tends to infinity?

As long as it is an equivalent quantity, it can be replaced in the multiplication and division method of limit operation, which has nothing to do with the change of independent variable
1. It must be equivalent. Your question seems not quite right. Because when x tends to zero, the two quantities are equivalent, but when x tends to infinity, the two quantities are not equivalent in general, so they can't be replaced
2. Only in multiplication and division, that is, the equivalent quantity of factors can be replaced

If you look at the following formula, what can you find? 11 out of 18 = 1 out of 3 + 1 out of 6 + 1 out of 9, 13 out of 30 = 1 out of 5 + 1 out of 6 + 1 out of 15
Can you fill in the following formula completely? (it must be a fraction, 1 / XX in brackets)
13 out of 24 = () + () + ()
15 out of 42 = () + () + ()
15 out of 36 = () + () + ()
13 out of 28 = () + () + ()
This is the 13th question on page 47 of the first volume of the sixth grade of primary school. Please teach me

1.3:1 + 8:1 + 12:1
1 / 2.6 + 1 / 7 + 1 / 21
1 / 3.4 + 1 / 9 + 1 / 18
1 / 4 + 1 / 7 + 1 / 14
This kind of problem is that the multiplication of the first two numbers is equal to the denominator number of the problem, and the third number is half of the denominator number of the current problem

Given that the fourth power of x plus the third power of MX plus NX minus 16 has the factors x minus 1 and x minus 2, find the value of M and N? And decompose the polynomial into factors

The so-called factorization is to decompose a polynomial into the form of multiplication of several factors, so the factorization after decomposition must be several terms of multiplication. The factorization (x-1) and (X-2) show that when x = 1 and x = 2 are substituted, the whole factorization is equal to 0, that is to say
1^4+m*1^3+n*1-16=0
2^4+m*2^3+n*2-16=0
So m = - 5, n = 20