A Nice Algebra Problem | Can you solve this ? | Math Olympiad

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Learncommunolizer

Learncommunolizer

Күн бұрын

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On How to solve for X in this Algebra, Maths Olympiad Problem
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Пікірлер: 20
@damirdukic
@damirdukic 18 күн бұрын
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.)
@yamunadegamboda8485
@yamunadegamboda8485 23 күн бұрын
With out using the formula can get the value of x by using factors
@key_board_x
@key_board_x 21 күн бұрын
@ 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
@user-ee7nw2rx9s
@user-ee7nw2rx9s 20 күн бұрын
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
@李梓豪-x2o
@李梓豪-x2o 22 күн бұрын
求根公式应该是2a/负b2+/-根号下b方-4ac。
@user-ee7nw2rx9s
@user-ee7nw2rx9s 20 күн бұрын
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
@HanhTran-tl8dm
@HanhTran-tl8dm 22 күн бұрын
Good solutions, thank you.
@CharlesChen-el4ot
@CharlesChen-el4ot 22 күн бұрын
(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
@robertocarniato6195
@robertocarniato6195 23 күн бұрын
b^-4 a c ....
@hansgodschalk466
@hansgodschalk466 23 күн бұрын
Indeed an error
@robertocarniato6195
@robertocarniato6195 23 күн бұрын
@@hansgodschalk466 io faccio di peggio
@user-de9vr5yq2h
@user-de9vr5yq2h 23 күн бұрын
Nice👍
@jyothsnapakanati5366
@jyothsnapakanati5366 22 күн бұрын
Good Job! 😀
@historyworldofarjun1870
@historyworldofarjun1870 23 күн бұрын
X = 1
@RyanLewis-Johnson-wq6xs
@RyanLewis-Johnson-wq6xs 22 күн бұрын
(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
@pspandey9737
@pspandey9737 23 күн бұрын
=>>>>> 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
@李梓豪-x2o
@李梓豪-x2o 22 күн бұрын
有一个步骤写错了
@user-xq8cg3gf5f
@user-xq8cg3gf5f 23 күн бұрын
х = 1 !!!
@JulioParrales-tk3iz
@JulioParrales-tk3iz 21 күн бұрын
VIDA SERES HUMANOS TIERRA 0
@RyanLewis-Johnson-wq6xs
@RyanLewis-Johnson-wq6xs 22 күн бұрын
x=±1 x=± 1+(Log[1.5,(32768/14348907)e^(2π)]+15)i final answer
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