6x-4x = 26

6x-4x = 26

4X ^ 2 - 4x + 1 = x ^ 2 + 6x + 9 is the best way to solve this equation
4X^2-4X+1=X^2+6X+9
(2x-1)^2=(x+3)^2
(2x-1)^2-(x+3)^2=0
(2x-1+x+3)(2x-1-x-3)=(3x+2)(x-4)=0
X = - 2 / 3 or 4
2X square + 1 = 3x (cross multiplication)
2x^2-3x+1=0
2 -1
1 -1
(2x-1)(x-1)=0
X = 1 / 2 or x = 1
I hope I can help you,
(2x-1) (x-1) = 0, so x = 1 / 2, x = 1
How to find general term formula in different sequence
Just as the relationship between an and an-1 is accumulated, the denominator is made up of the arithmetic sequence by the split term method, and the arithmetic sequence by the coefficient multiplication and the arithmetic sequence by the dislocation subtraction method (please be more complete)
Due to the word limit, I posted something to my blog. Some of the content was not finished, but it was enough to deal with common sequence problems
The solution set of 2x squared - 3x-2 > 0 is a process
2x² - 3x - 2 > 0
(2x + 1)(x - 2) > 0
X < - 1 / 2 or x > 2
2x² - 3x - 2 > 0
(2x + 1)(x - 2) > 0
X < - 1 / 2 or x > 2
General term formula of sequence
The odd term of the sequence constitutes the arithmetic sequence with tolerance of - 2, and the even term constitutes the arithmetic sequence with common ratio of 2, A1 = 12, A2 = 2, to find an
an=(13-n)*[0.5*(-1)^(n-1)+0.5]+2^(n/2)*[0.5*(-1)^n+0.5]
Square of X - 3x + 2 cross multiplication
X -1
X -2
The formula of finding general term in sequence
-1,1,5,11,19?
A:
A1=-1
A2=1,A2-A1=2
A3=5,A3-A2=4
A4=11,A4-A3=6
A5=19,A5-A4=8
So:
A(n+1)-An=2n
Add the above formulas:
A(n+1)-A1=2*(1+n)n/2
A(n+1)+1=n^2+n
An=n(n-1)-1
-1+2=1
1+4=5
5+6=11
11+8=19
an-a(n-1)=2(n-1)
……
a2-a1=2*1
Add
an-a1=2*[1+2+…… +(n-1)=n(n-1)
a1=-1
So an = n & # 178; - n-1
An=n²-n-1
-1, 1, 5, 11, 19... To find the general term formula?
a₁=-1，a₂=1，a₃=5，a₄=11，a₅=19，...........；a‹n›=?
a₂-a₁=1-(-1)=2
a₃-a₂=5-1=4
A &; - A &; =... Expand
-1, 1, 5, 11, 19... To find the general term formula?
a₁=-1，a₂=1，a₃=5，a₄=11，a₅=19，...........；a‹n›=?
a₂-a₁=1-(-1)=2
a₃-a₂=5-1=4
a₄-a₃=11-5=6
a₅-a₄=19-11=8
..................
a‹n›-a‹n-1›=2+2[(n-1)-1]=2n-2
__________________________ +
a‹n›-a₁=2+4+6+8+......+(2n-2)=[2+(2n-2)](n-1)/2=n(n-1)
So a &; n &; = A &; + n (n-1) = n &; - n-1
This is the general formula. Put it away
How to solve x-square-3x-28 = 0 by cross phase multiplication
(x+4)(x-7)=0
X = - 4 or x = 7
x^2-3x-28=0
1 -7
1 4
=-7+4=-3
(x-7)(x+4)=0
x=7 x=-4
x^2-3x-28=0
1 -7
1 4
=-7+4=-3
(x-7)(x+4)=0
x=7 ，x=-4
I hope you can understand it. I wish you progress in your study
1 4
1 -7
If you want to add before and after crossing, cross and multiply
(X-7) * (x + 4) = 0: x-6x + 8 = 0
How to find the general term formula of sequence
Li Weiguo, the third senior high school in Jianchang County, Liaoning Province
1、 Summation (also called summation by difference)
2、 Cumulative multiplication (also called quotient quadrature) 4. Formula method
There are two main points to be noted in the reflection of reciprocal transformation: one is to take the reciprocal; the other is to pay attention to the first term of the new sequence, and the tolerance or common ratio has changed
6、 Undetermined coefficient method
In addition, there are logarithmic transformation method, iterative method, induction method, substitution method and so on