I divided both the numerator and the denominator with "8^x", then I introduced the substitution "u = (3/2)^x", thus obtaining the equation "6u^3 - 7u^2 - 7u + 6 = 0". "u = -1" is the most obvious solution. After we divide this equation with "u + 1", we obtain the equation "6u^2 - 13u + 6 = 0", which has two solutions: 2/3 and 3/2. So, "u" can be 1, 2/3 or 3/2 and since u is (3/2)^x, x can be -1 or 1. (If we search the solution in the real number set.)
Поделить на 8^х (1+(3/2)^(3х))/((3/2)^(х)+(3/2)^(2х))=7/6 Легко видно что а=(3/2)^х (1+а^3)/(а^2+а)=7/6 В итоге 6а^2-13а+6=0 Легко решить И никаких лишних замен и двух переменных не нужно
Вы так легко сократили на а+b, а если a+b=0, что тогда? (a/b) =( - b+sqrt (b^2-4ac)/2a Почему-то слева b= 13, a=6 Тогда справа 6/13=3/2, что в корне не верно
Good solutions, thank you.
I divided both the numerator and the denominator with "8^x", then I introduced the substitution "u = (3/2)^x", thus obtaining the equation "6u^3 - 7u^2 - 7u + 6 = 0". "u = -1" is the most obvious solution. After we divide this equation with "u + 1", we obtain the equation "6u^2 - 13u + 6 = 0", which has two solutions: 2/3 and 3/2. So, "u" can be 1, 2/3 or 3/2 and since u is (3/2)^x, x can be -1 or 1. (If we search the solution in the real number set.)
@ 4:55 / 11:54 → do not make cross multiply, but divide by ab (directly)
(a² - ab + b²) / ab = 7/6
(a²/ab) - (ab/ab) + (b²/ab) = 7/6
(a/b) - 1 + (b/a) = 7/6
(a/b) - 1 + [1/(a/b)] = 7/6
(a/b) + [1/(a/b)] = (7/6) + 1 → let: m = (a/b)
m + (1/m) = 13/6
m² + 1 = (13/6).m
m² - (13/6).m + 1 = 0
Δ = (13/6)² - 4 = (169/36) - (144/36) = 25/36 = (5/6)²
m = [(13/6) ± (5/6)]/2
m = (13 ± 5)/12
m = (13 + 5)/12 = 18/12 = 3/2
m = (13 - 5)/12 = 8/12 = 2/3
Recall: m = a/b = 2^(x)/3^(x) = (2/3)^(x)
First case: m = 3/2
(2/3)^(x) = 3/2
(2/3)^(x) = [1/(2/3)]
(2/3)^(x) = (2/3)^(- 1)
→ x = - 1
Second case: m = 2/3
(2/3)^(x) = 2/3
(2/3)^(x) = (2/3)^(1)
→ x = 1
Good Job! 😀
With out using the formula can get the value of x by using factors
Поделить на 8^х
(1+(3/2)^(3х))/((3/2)^(х)+(3/2)^(2х))=7/6
Легко видно что а=(3/2)^х
(1+а^3)/(а^2+а)=7/6
В итоге
6а^2-13а+6=0
Легко решить
И никаких лишних замен и двух переменных не нужно
求根公式应该是2a/负b2+/-根号下b方-4ac。
Nice👍
Вы так легко сократили на а+b, а если a+b=0, что тогда?
(a/b) =( - b+sqrt (b^2-4ac)/2a
Почему-то слева b= 13, a=6
Тогда справа 6/13=3/2, что в корне не верно
=>>>>>
6m^2-13m+6=0
6m^2-9m-4m+6=0
3m-2)(2m-3)=0.....
.(2/3)^×=2/3,,,,,=>×=1
2m=3.....m=2/3)^×
=>×=-1
b^-4 a c ....
Indeed an error
@@hansgodschalk466 io faccio di peggio
（2^x + 3^x)(2^2x + 3^2x+ 2^x*3^x)/
6^x* ( 2^x+3^x) = 7/6
6*(2^2x + 3^2x + 2^x*3^x) = 7*6^x
6*(2^2x+3^2x)= 6^x
(8^x+27^x)/(12^x+18^x)=7/6 x=±1 x=± 1+(Log[1.5,(32768/14348907)e^(2kπ)]+15)i k=z
X = 1
有一个步骤写错了
х = 1 !!!
VIDA SERES HUMANOS TIERRA 0
x=±1 x=± 1+(Log[1.5,(32768/14348907)e^(2π)]+15)i final answer