Пікірлер

• 
  @davidshen59162 күн бұрын

  X^4=（1-2X）^2=1-4X+4X^2=1-2（1-X^2）+4X^2=6X^2-1， X^4+6/X^2=6（X^2+1/X^2）-1=6（X-1/X）^2+12-1=6（-2）^2+11=35

• 
  @sksharma8263 күн бұрын

  what is 1V and Vo

• 
  @sharonmarshall36717 күн бұрын

  I reduced 87/21 to 3 from the beginning which reduced the working going forward

• 
  @stephenhall56948 күн бұрын

  Left out the square (power of 2) mark on the intermediate step, just before the quadratic equation step

• 
  @SidneiMV9 күн бұрын

  x^x⁴ = 64 = 8² x⁴^x⁴ = 8⁸ x⁴ = 8 => *x = 8^(1/4)*

• 
  @SidneiMV10 күн бұрын

  (x² + 1 + 1/x²)/(x + 1 + 1/x) = 3 x² + 1 + 1/x² = 3(x + 1 + 1/x) (x + 1/x)² - 3(x + 1/x) - 4 = 0 x + 1/x = u u² - 3u - 4 = 0 (u - 4)(u + 1) = 0 u - 4 = 0 => u = 4 x + 1/x = 4 x² - 4x + 1 = 0 x = (4 ± 2√3)/2 *x = 2 ± √3* u + 1 = 0 => u = -1 x + 1/x = -1 x² + x + 1 = 0 *x = (-1 ± i√3)/2*

• 
  @user-xh3ih4ks9y14 күн бұрын

  X=3. Y=7 故 X^2+Y^2=58

• 
  @pas629514 күн бұрын

  58

• 
  @MathEducation100M16 күн бұрын

  Nice trick

• 
  @Naduni349518 күн бұрын

  X^2 + Y^2 = ? X+Y= 10 XY=21 (a+b)^2 = a^2 +2ab + b^2 (x+y)^2=x^2+2xy+y^2 10^2=x^2+y^2+(2 . 21) 100=x^2+y^2+42 x^2+y^2=100-42 x^2+y^2=58

• 
  @neelamsaini663420 күн бұрын

  ❤😊😊

• 
  @user-ee7nw2rx9s21 күн бұрын

  Можно идти в обратную сторону 81=9^2, тогда Sqrt (3^x*(sqrt 9^x*(sqrt (9^x)^2=sqrt (3^x*sqrt( (9^x)^2))=sqrt (3^x*9^x)=sqrt( (3*9)^x=sqrt (3^(3*x)=3^(3*x/2)=3^4 3x/2=4, x=8/3 Кстати это уже 5 человек который решает этот пример

• 
  @ramieskola784529 күн бұрын

  Your second line is incorrect.

• 
  @ramieskola784529 күн бұрын

  You missapplied the exponent rule twice to land to the correct answer.

• 
  @Mb-logic28 күн бұрын

  I also make these type of videos​@@ramieskola7845

• 
  @ashrafsafwat3688Ай бұрын

  Very nice.

• 
  @amir_ismailiАй бұрын

  interesting 👏👏

• 
  @Math_with_aram24 күн бұрын

  Glad you think so!

• 
  @rezaghaderpour8725Ай бұрын

  👌👌👌

• 
  @mohinkhan2503Ай бұрын

  Excellent

• 
  @benjaminvatovez8823Ай бұрын

  Thank you for the video. I don't understand why you supposed that a,b are both even when they are clearly not as (a^b-b^a) is odd.

• 
  @BR-lx7pyАй бұрын

  @2:00 why are x and y necessarily integers?

• 
  @Math_with_aramАй бұрын

  This is just a placeholder for less writing and its not necessary

• 
  @Math_with_aramАй бұрын

  The paper becomes less crowded

• 
  @calamitates77Ай бұрын

  Isn't a = 18, b = 1 a (very obvious) solution too?

• 
  @haydencook9298Ай бұрын

  there are infinitely many solutions, given that a,b are real numbers

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  @benjaminvatovez8823Ай бұрын

  @@haydencook9298 So why has this been solved like they were integers?

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  @haydencook9298Ай бұрын

  @@benjaminvatovez8823 They just wanted a specific solution, not general solution

• 
  @user-je4wg1ow6oАй бұрын

  Great❤

• 
  @Math_with_aramАй бұрын

  Thanks

• 
  @mab9316Ай бұрын

  Beautiful

• 
  @Math_with_aramАй бұрын

  Thank you!

• 
  @ambassadorkeesАй бұрын

  I don't see it proven that this is the only solution. After all, even with this answer, there's a 3rd and 4th power in the problem. x-y>0 doesn't require both being positive and x>y, it can also be both negative and x<y. That's a path to and set of solutions.

• 
  @sonicwaveinfinitymiddwelle85552 ай бұрын

  lol

• 
  @Math_with_aram2 ай бұрын

  Why

• 
  @A.-us6jl2 ай бұрын

  thanks bor

• 
  @ceciliafreitas78562 ай бұрын

  15^30

• 
  @mediaguardian2 ай бұрын

  The answer is 15^30. You can't cancel exponentiation the same way you cancel divisors.

• 
  @nalinivijayan56172 ай бұрын

  My approach is like this 30^60/60^30 , here we can note that power of 30 is n×2 the power of 60 , so on that case 30^2/60^1 = 900/60 = 15 So if it was 30^60 / 60^30 then it will be eqaul to 15^30 as 30^2 / 60^1 = 15 , also 30^4 / 60^2 = 15^2 hence 30^60/60^30 = 15^30

• 
  @astrosamurai71352 ай бұрын

  or you could 30th root the whole thing, leaving you with 30^2/60, then simplify 90/60 to 3/2

• 
  @astrosamurai71352 ай бұрын

  two steps, much better than the 10 or something shown here.

• 
  @Math_with_aram2 ай бұрын

  Great solution yes

• 
  @letitburn33272 ай бұрын

  You can't do that when it's a simplification problem and not an equation, you could write it as the 30th root of everything to the 30th power so you would get (30^2/60)^30 leaving you with the same solution as the video

• 
  @TamirKorem2 ай бұрын

  This is inaccurate: There are 4 solutions to this exercise and not only two...: Two of them are real: 1) X+Y = 9 and 2) X+Y=-9 and two of them are complex (imaginary): 3) X+Y = 3i 4) X+Y= - 3i i = is the square root of minus 1

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  @HUBZONE-19Ай бұрын

  leave imaginery numbers alone 😵

• 
  @ksmyth9992 ай бұрын

  (x+y)(x-y) = 27. This suggests x+y = 9, x-y =3. So 2x = 12, x = 6, y =3

• 
  @TamirKorem2 ай бұрын

  You cannot rely on a guess... There are 4 solutions to this exercise and you will miss three of them by using your guess. See my response in the other comments.

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  @ksmyth9992 ай бұрын

  @@TamirKorem Thanks for the comment on my comment. Yes there are four solutions. But my suggestion was not a wild guess, it was a logical conclusion on the assumption one of the solutions would be real integers. That there might be real integer solutions was admittedly a guess, but it was easily testable. The idea of this exercise, in my opinion, was to find one of the roots using the difference of two squares. It was something the presenter had missed. The presenter went on to find the two real solutions. It was good that you pointed out the two complex solutions.

• 
  @JuPiTeR_02112 ай бұрын

  great sir😀

• 
  @ffoo93842 ай бұрын

  seriously, math olympiad question? first desicion (without math at all) multiplying x*y = -15 it means first (x or y) is with - second (x or y) with + x-y = 8, means x greater than y, so x with + sign y with - sign multiplying gives us nonpair integer and dividing gives us pair integer it means that x is positive nonpair integer y is negative nonpair integer not so hard to solve x = 5 y = -3 5^4 +(-3)^4 = 706 second desicion x - y = 8 => x=8+y y*(8+y) + 15 = 0 y^2 +8*y +15 = 0 y = -3 x=8+y = 5 5^4 +(-3)^4 = 706 why the hell we are playing with formulas when answer is so obvious?

• 
  @pedroalves24122 ай бұрын

  brilliant

• 
  @mohinkhan25032 ай бұрын

  1?

• 
  @franko.w.27422 ай бұрын

  Thx 😎👍

• 
  @maxence_lvl83312 ай бұрын

  Hi! I don't quite understand... I achieved to isolate the x's with a polynomial which equals to 0 (--> 48x^2-16x+1=0). And i found two roots : 1/4 and 1/12. Of course 1/12 doesn't work, but I don't understand why. Could you explain this to me please ? Thanks

• 
  @lightyagami17522 ай бұрын

  Substitute it back into the original equation and you'll find it does not satisfy it. This is a known phenomenon when you take squares or other even higher powers of equations, you may introduce redundant or extraneous roots. This is because by standard convention, sqrt(x) with the usual symbol, only refers to the non-negative square root. But when you square you unwittingly introduce the possibility of including the negative square root. That's why you need to check every single solution by substitution into the original equation.

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  @maxence_lvl83312 ай бұрын

  @@lightyagami1752 Thanks for the response!

• 
  @tejasvadhyani78602 ай бұрын

  @@lightyagami1752 I think 1/12 is a valid value since sqrt(25) = -5 as well. Since the degree is 2, we would get 2 zeroes(values)

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  @user-sb5ql4qz1f2 ай бұрын

  the solution was damn good

• 
  @user-sp7qz4ur4r2 ай бұрын

  missed 2nd root, if t=-5 x=1/12

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  @skill_issuesmo73672 ай бұрын

  t cant be negative

• 
  @kiflakiflakifla2 ай бұрын

  If you cannot notice such elegant polynomial transformations and factorizations. Let f(x) = x⁹+x⁶-36 => f'(x) = x⁵(9x³+6) => f'(x) < 0 <=> x is in interval (-⅔^⅓, 0) => for all x < 0 f(x) <= f(-⅔^⅓) < 0 (since the function is increasing until -⅔^⅓ and then decreasing afterward until 0.) => there are no solutions for x < 0. Notice that x = 0 isnt a solution. If x > 0 then f'(x) > 0 which implies there can be at most one solution (if the function is always increasing it can intersect x axis at most once) By playing around you can see that 3³+3² = 36. (i first tried out 1, and then 2, and then 3...) Hence x³ = 3 <=> x= 3^⅓ is the only real solution.

• 
  @samuelbudzinak2 ай бұрын

  4:00, I would preffer "no real solution"

• 
  @ryanpethick6502 ай бұрын

  When the discriminant is less than 0 it doesn’t mean it’s impossible it just means that there’s no real solutions. You still have imaginary solutions

• 
  @Math_with_aram2 ай бұрын

  sure, thanks for your inforamtions

• 
  @user-uc2il1sg9s2 ай бұрын

  It's amazing how easy it is to solve, but at the same time how difficult it is

• 
  @Math_with_aram2 ай бұрын

  sure bro❤

• 
  @leontheeuwen10502 ай бұрын

  Dahm

• 
  @gabrielandre75492 ай бұрын

  Don’t the exponent add up to 6? Since you multiplied everything

• 
  @astro88332 ай бұрын

  No

• 
  @VividBoricua2 ай бұрын

  (x^4)(x^2)(x) = (x^4)(x^2)(x^1) That last x has an implied exponent value of 1, not 0.

• 
  @Scotty-Tea2 ай бұрын

  Helpful vid, thanks Aram!

• 
  @Math_with_aram2 ай бұрын

  Thanks for you attention ❤❤

• 
  @amir_ismaili2 ай бұрын

  interesting 👏👏👏

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  @Math_with_aram2 ай бұрын

  Thanks for watching