If the equation (M & sup2; - 1) x & sup2; - XM + 8 = x is a linear equation with one variable, then the value of the algebraic formula m ^ 2008 - | M-1 | is

If the equation (M & sup2; - 1) x & sup2; - XM + 8 = x is a linear equation with one variable, then the value of the algebraic formula m ^ 2008 - | M-1 | is

(m²-1)x²-xm+8=x
(m²-1)x²-(m+1)x+8=0
There is no quadratic term for the linear equation of one variable x
So the quadratic coefficient is 0
m²-1=0
m²=1
M = 1 or - 1
And the first term is not equal to 0
m+1≠0
So m = 1
therefore
m^2008-|m-1|=1^2008-|1-1|=1-0=1
(m²-1)x²-xm+8=x
(M & sup2; - 1) x & sup2; - x (M + 1) + 8 = 0 is a linear equation with one variable
M ^ 2-1 = 0, and M + 1 ≠ 0
So, M = 1
m^2008-|m-1|=1^2008-|1-1|=1-0=1
(m²-1)x²-xm+8=x
（m²-1)x²-(m+1)x+8=0
Because it is a linear equation of one variable, the quadratic term is 0, so m = 1 or - 1. Moreover, the quadratic term cannot be 0, so m is 1
m^2008-|m-1|=1-0=1
The original equation can be reduced to (M & sup2; - 1) x & sup2; - (M + 1) x + 8 = 0. From the meaning of the problem, we know that the coefficient of quadratic term M & sup2; - 1 = 0, and the coefficient of primary term m + 1 ≠ 0. The solution is m = 1
So the algebraic formula m ^ 2008 - | M-1 | = 1-0 = 1
A problem of merging similar terms and transferring terms in linear equation of one variable
The distance between station a and station B is 360km, with an express train running 72km per hour from station B and a slow train running 48km per hour from station a
(1) How many hours has it taken for the two trains to meet each other?
(2) The express train leaves for 25 minutes first, and the two trains are facing each other. How many hours has the local train been running?
1. Let's meet after X hours
Same as the first question
72X+48X=360
120X=360
X=3
2. Set the local train to run for X hours, and the two cars meet
When the two cars meet, they travel the whole distance
72(X+25/60)+48X=360
3(X+25/60)+2X=15
3X+75/60+2X=15
X=15-75/60
X=13.75
3x-4 times 5 = 16 solve the following equation
Does it mean 3x-4x5 = 16?
So simplify first: 3x-20 = 16
Transfer: 3x = 16 + 20
Merge similar items: 3x = 36
Divide both sides of the equation by 3 to get x = 12
.
The solution of the original equation is x = 12
3x-4*5=16
3x-20=16
3x=36
x=12
6x-7=4x-5．
6x-7 = 4x-5, transfer term, get: 6x-4x = - 5 + 7, merge similar term, get: 2x = 2, coefficient is 1, get: x = 1
2X minus (8 times 0.9) product, equal to 2.4 3x plus (1.5 times 1.2), equal to 16.8 equation
The product of 2x minus (8 times 0.9) equals 2.4
2x-7.2=2.4
2x=2.4+7.2
2x=9.6
x=4.8
3x plus (1.5 times 1.2) equals 16.8
3x+1.8=16.8
3x=16.8-1.8
3x=15
X=5
2x-0.72=2.4
x=1.56
3x+1.8=16.8
3x=16.8-1.8
X=5
Request equation 4x + 48 = 6x-8
48+8=6x-4x
56=2x
x=28
4x+48=6x-8
2x=56
x=28
4x+48=6x-8
2x=56
x=28
4x+48=6x - 8
4x - 6x=-48-8
-2x=-56
x=28
I want to know how to do the equation of 3x-4 × 5 = 16?
3x-20=16
3x=16+20
3x=36
x=12
No, x = 12
What is the solution of equation 6x-26 = 18 + 4x
6x-4x=26+18
2x=44
x=44÷2
x=22
6x-26=18+4x
6x-4x=18+26
2x=44
x=22
6x-4x=18+26
2x=44
x=22
How to solve the equations 3x-10 equals 80, 25% x + x equals 10, x + 160% x equals 65
3x-10=80
3x=90
x=30
25%+x=10
1.25x=10
X=8
x+160%x=65
2.6x=65
x=25
3X-10=80 3x=90 x=30
25%X+X=10 1.25x=10 x=8
X+160%X=65 2.6x=65 x=25
3x-10=80 → x=(80+10)/3=30
25%x+x=10 → (0.25+1)x=10 → x=10/1.25=8
x+160%x=65 → (1+1.6)x=65 → x=65/2.6=25
1. 3x-10 = 80 --- move to -- 3x = 90 --- x = 30
2. X / 4 + x = 10 --- multiply by 4 --- 5x = 40 --- x = 8 at the same time
3. X + 1.6x = 65 --- multiply by 10 at the same time --- 26x = 650 --- x = 25
3x-10=80
3x=80+10
3x=90
X=90/3
X=30
5.6x = 18 minus 4.4x 8 / 5 divided by (x minus 0.45) equals 16 / 5 how to solve these two equations? My equation is not good, please help me
5.6x = 18 minus 4.4x
5.6x+4.4x=18
10x=18
x=18÷10
x=1.8
8 out of 5 divided by (x minus 0.45) equals 16 out of 5
x-0.45=8/5÷16/5
x-0.45=0.5
x=0.45+0.5
x=0.95