How to solve the equation 6.2-4x + 6x = 10.8

How to solve the equation 6.2-4x + 6x = 10.8

6.2-4x+6x=10.8
6.2+2x=10.8
2x=10.8-6.2
2x=4.6
x=2.3
6.2-4x+6x=10.8
6.2+2x=10.8
2x=10.8-6.2
2x=4.6
x=2.3
2x=4.6 x=2.3
6x-4x = 26, the equation is completely solved
Er, er, mistake
For the function f (x), X ∈ D, a sequence generator can be constructed. Its working principle is as follows: (1) input x0 ∈ D, input X1 = f (x0) through the sequence generator; (2) if X1 does not belong to D, the generator will end. If x1 ∈ D, X1 will feed back to the input, and then input x2 = f (x1), and continue in turn
Now define f (x) = (4x-2) / (x + 1)
First question: if you want the sequence generator to generate an infinite constant sequence, try to find the value of the input initial data x0
Second question: if the infinite sequence {xn} generated by inputting x0 satisfies: xn exists for any positive integer n
Let f (x) = x, x = 2 or 1
2. Let f (x) > x, get 1 ＜ x ＜ 2, or X ＜ - 1. After verification, if 1 ＜ x ＜ 2, then 1 ＜ f (x) ＜ 2, so that all sequences are satisfied. If x ＜ - 1, then f (x) ＞ 2, then f (f (x)) ＜ f (x), not satisfied, discard ~ ~ to sum up, 1 ＜ x0 ＜ 2
How to solve the equation (x-3) + (4x-3) = 49
Good!
（x-3）+（4x-3）=49
5x-6=49
5x=55
x=11
（x-3）+（4x-3）=49
x-3+4x-3=49
x+4x=49+3+3
5x=55
x=55/5
x=11
Remove bracket 5x-6 = 49
5x=55
x=11
The solution of the general term formula of the sequence of difference into equal difference
What is the general formula of sequence 1,2,4,7,11,16,22? What is the algorithm of similar formula?
an=a(n-1)+n-1
a(n-1)=a(n-2)+n-2
.........
a2=a1+1
an=a1+1+2+...+(n-1)
=1+n(n-1)/2
Recursion
an=(n^2-n+2)/2
It is solved by accumulation. The latter term is subtracted from the former term, and then accumulated
How to solve this equation? 24-4x = 2x-6
Move item 24 + 6 = 2x + 4x
30=6x
X=5
24-4x=2x-6
-4x-2x=-6-24
-6x=-30
X=5
5 = 5 = 5 = 5 = 5 = 5 = 5 = 5...
The solution of general term formula of sequence
1. Accumulation
Given that the sequence {an} satisfies an + 1 = an + 2n + 1, A1 = 1, find an
2. Cumulative multiplication
Given that the sequence {an} satisfies A1 = 2 / 3, an + 1 = n / (n + 1) an, find an
3. Construct new sequence
In known sequence {an}, A1 = 1, an = 2an-1 + 1 (n ≥ 2), find an
Note: n - + 1 in an + 1 or an-1 is a diagonal sign
1 accumulation because a (n + 1) = = an + 2n + 1, so an = a (n-1) + 2 (n-1) + 1. (1) a (n-1) = a (n-2) + 2 (n-2) + 1. (2) a (n-2) = a (n-3) + 2 (n-3) + 1. (3)... A2 = a1 + 2 * 1 + 1. (n-1) A1 = 1. (n) the square of an = n + 2 {(n-1) + (n-2) + (n-3) +. + 1} = n is obtained from (1) to (n)
a(n+1)=an+2n+1
a(n+1)-an=2n+1
an-a(n-1)=2n-2+1
.....
a2-a1=2+1
Add a (n + 1) = a1 +
(on the right is the summation of arithmetic sequence, and the formula is 2n + 3)
In the same way, write out the items before and after each other
3 add a number m to the left so that an + M = 2 (an-1 + m)
So m = (M + 1) / 2... Expansion
a(n+1)=an+2n+1
a(n+1)-an=2n+1
an-a(n-1)=2n-2+1
.....
a2-a1=2+1
Add a (n + 1) = a1 +
(on the right is the summation of arithmetic sequence, and the formula is 2n + 3)
In the same way, write out the items before and after each other
3 add a number m to the left so that an + M = 2 (an-1 + m)
So m = (M + 1) / 2 (because there is a constant 1 on the right side of the equation, adding a number on both sides of the equation still holds, and the coefficient 2 of an-1 is put forward)
m=-2
Take An-2 as a whole, that is, a new sequence BN
BN is an equal ratio sequence
Find out BN and put it away in finding an
How to solve the equation 2x-9 + 4x = 6
2X-9+4X=6
6x=6+9
x=5/2
The solution of general term formula of sequence
Should the general term formula for a (n) be a (n) = s (n) – s (n – 1),
It needs to be verified because the first term may not conform to the general term formula
(1) 4,2,3,4,…… N
(2) 3,4,8,…… 2^n
N>1
If x = 4, y = 1 are the common solutions of the equations 4x + my = 9 and MX NY = 11, then M=__ ,n=__
Taking x = 4 and y = 1 into the first equation can get m = 7, and then taking these three values into the second equation can get n = 17
m=7 n=17
Take it in and you'll know····