Пікірлер

• 
  @Newedomaya_Hren2 күн бұрын

  "Что-то умное про подбор"

• 
  @marcofogs3703Ай бұрын

  This is the righ way to solve such question. Of course in this example is easy to find 3 as solution. But in general case , this is the right way to solve, mainly when the the solution is not a integer number.

• 
  @marcofogs3703Ай бұрын

  perfect.....

• 
  @user-ec5ip3vp2rАй бұрын

  (-9;8)(8;-9)

• 
  @ranshen14862 ай бұрын

  Why not just replace -y by x+1 in eqn (1) once we get x+y+1=0?

• 
  @b33017450162 ай бұрын

  Set x=y-3. It will be easier to solve y.

• 
  @greatindianbull2 ай бұрын

  Correct

• 
  @valeriocolombini39002 ай бұрын

  Hi, can you solve x³+x²=27 ? Thank you!!

• 
  @jamiewalker3292 ай бұрын

  The mean/median of x, x+2, x+4 and x+6 is x+3: so to restore some symmetry i'd be tempted to let u = x+3 from the outset so that the equation becomes (u-3)(u-1)(u+1)(u+3) = 9. Hence (u^2 - 1)(u^2-9) = 9 so u^4 - 10u^2 = 0 so u^2( u^2 - 10) = 0. This gives u^2 = 0 or 10 or equivalently u = 0 or ±sqrt(10) and so x = -3, -3 ±sqrt(10)

• 
  @rdan64632 ай бұрын

  👍 thank you

• 
  @chavaamu2 ай бұрын

  1/4

• 
  @mathsdaily1842 ай бұрын

  X = -3

• 
  @davidbrisbane72062 ай бұрын

  Let x = u - 3 and let's find the integer solutions (if they exist). So, (u - 3)((u - 1(u + 1)(u + 3) = 9 So, (u³ - 1)(u² - 9) = 9 So, u² is 0, so u is 0, as u² - 1 and u² - 9 are factors of 9. So x = - 3. In fact, this is a double root of the original formula and we are left to find the other two solution, which are non-integer but could be complex solution, since the original equation is a quartic equation. So we could expand the original equation and divide it by (x + 3)² to find the remaining quadratic formular and then solve it using the quadratic formula.

• 
  @benovision63252 ай бұрын

  x=-3

• 
  @DandoPorsaco-ho1zs2 ай бұрын

  You are being fooled with this nonsense! This method only works if you know the solution in advance. You need to split a number like 30 into two numbers x and y, such that x + y = 30 and 3^x = y (ie. 3 +27 = 30 and 3^x = 27), where x must be the solution you are trying to calculate. In other words, you need the solution of the problem in order to successfully follow the last steps. For example, if you try to solve 3^x + x = 40, you just need to split 40 into 3.279887… (the solution to the exercise!) and 36.72011..., both irrational numbers (infinite decimals), since 3.279887… + 36.72011... = 40 and 3^3.279887… = 36.72011... Brilliant method!

• 
  @alejomdp2 ай бұрын

  Have you tried your solutions into the original equation? I can't make them work. I think you made a mistake when cancelling squares with square roots. To me the problem has no solution: √x + √(-x) = 2 √x + √((-1)x) = 2 √x + i √x = 2 Here, the real parts on both sides of the equation have to be the same, and also the imaginary parts have to be the same. That's not possible, so there are no solutions.

• 
  @HijaziSpeaks2 ай бұрын

  Solved beautifully

• 
  @HijaziSpeaks2 ай бұрын

  Solved nicely sir

• 
  @rovicmartecio24403 ай бұрын

  Hi! I only factored the expression to solve it but it's nice to see your excellent algebraic solution which can be a tool for me/everyone to solve further mathematical problems. Thanks!

• 
  @HijaziNetwork3 ай бұрын

  Nice explanation

• 
  @avijitlahiri59023 ай бұрын

  good approach

• 
  @doordie25433 ай бұрын

  Easy

• 
  @doordie25433 ай бұрын

  Very easy

• 
  @abhirupkundu27783 ай бұрын

  If you cannot speak english like a native american, do not try to, you keep making a fool of yourself.

• 
  @HijaziNetwork3 ай бұрын

  Very intresting

• 
  @jackrobbins093 ай бұрын

  Excellent solution

• 
  @worstcase10763 ай бұрын

  There is noting to solve here, this is already a number, not an equation.

• 
  @Smart_Root3 ай бұрын

  2:21 : I left the chat

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  @user-rj8oj6tk8l3 ай бұрын

  Nice❤

• 
  @akramabdou30053 ай бұрын

  not correct 255/16

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  @user-rj8oj6tk8l3 ай бұрын

  I am about to comment the same

• 
  @akramabdou30053 ай бұрын

  255/16

• 
  @1mthoughts4573 ай бұрын

  Solved it smoothly

• 
  @user-su3km4bl9e3 ай бұрын

  As a generalization, the method presented always works if in the equation a^x + x = b one has b = a^a + a, with the solution x = a. Try it out with some numbers or repeat the derivation as in the video.

• 
  @usmanwyne85034 ай бұрын

  sometimes there is no need to think a lot. the first thought after looking is correct answer so no need to over complicate.

• 
  @user-sx6vo6ne5f4 ай бұрын

  ...chłopie, po co tyle pierdolenia ?!...a+b=-1 -> a=-b-1 wstawiasz do drugiego równania i masz równanie kwadratowe z b -> { (-9,8); (8,-9)}

• 
  @fisicamatematicasprofewilliam4 ай бұрын

  Nuevo suscriptor. gran problema de algebra, apoyo mutuo. like

• 
  @hajeo44615 ай бұрын

  нахуя а главное зачем

• 
  @Bellmak5 ай бұрын

  Ты крут, люди считающие, что ты всё усложнил просто смешны. Да, здесь x легко находится подбором, но посмотрим на них, когда в следующий раз x окажется не целым числом, например.

• 
  @razetqa5 ай бұрын

  Биномиальный поиск

• 
  @19Dima985 ай бұрын

  да и плюс когда решаешь алгебраически, то находишь все корни уравнения или (как здесь) доказываешь, что корень только один. решая подбором ты никогда не можешь быть уверен, что у уравнения не может быть ещё какого-то корня, который ты не учёл. подбор - всегда неполное решение, т.к. ты не доказываешь, что больше корней нет. геометрическое решение тоже не всегда верно, т.к. оно показывает пересечение графиков функций на действительной плоскости, а как быть уверенным, что у уравнения нет ещё дополнительного комплексного корня? его не изобразить геометрически, почти невозможно найти подбором, а решая такое уравнение алгебраически, ты всегда на 100% находишь все корни и никто не подкопается

• 
  @sunnatradjapov10054 ай бұрын

  Всё супер

• 
  @DandoPorsaco-ho1zs2 ай бұрын

  It doesn't even have to be an integer! The split 30 = 3 + 30 requires the solution (eg. 3) that you are looking for. If you don't know the solution, you can't use this method.

• 
  @KonstantinKostin6232 ай бұрын

  Сумма двух возрастающих функций - возрастающая функция, так что тут очевидно только одно решение. Так что это явное переусложнение. Если бы ответ был нерациональным чистом, тогда и стоило бы так решать

• 
  @Ivanofficial525 ай бұрын

  Bro doesn't search for easy ways

• 
  @Ivanofficial523 ай бұрын

  Actually when I wrote this I had no idea how to solve this and neither I do now. Sorry

• 
  @thatonenoname3 ай бұрын

  The WORKING easy way is make graphs of functions 3^x and 30 - x. With this graph we can prove that this equation has only one solution. People can just understand it's 3, but robots will use perfect made graph for it

• 
  @dwiyantowibisana49285 ай бұрын

  x=y=10

• 
  @alimehmani68686 ай бұрын

  b= -9 a=+8 and vise vers.

• 
  @MurtazaAhmed-mq1bh6 ай бұрын

  ❤❤❤❤

• 
  @mastisemastishktak57506 ай бұрын

  bro u can do it easily by binomial expansion of (1+.005)^200 its first and second term are both 1 and we are done

• 
  @MurtazaAhmed-mq1bh6 ай бұрын

  ❤❤❤❤

• 
  @thanhbui2286 ай бұрын

  Excellent!

• 
  @LetsAssume_6 ай бұрын

  Thank you ❤️

• 
  @AA-Logic6 ай бұрын

  Very nice ❤

• 
  @LetsAssume_6 ай бұрын

  Thanks 🤗

• 
  @AA-Logic6 ай бұрын

  Thank you

• 
  @AndyU966 ай бұрын

  Not bad, but maybe change the base from 2 to something else if it's even possible to do so. This was too easy to solve by thinking alone, 2048 - 16

• 
  @roger73416 ай бұрын

  11^33*2^33 vs 11^22*3^22 Refactor Divide by 11^22 and 2^22 yields 11^11*2^11 vs (3/2)*22 22^11 vs (9/4)^11 Change base and exponent and compare bases 22 vs (9/4) so 22^33 is larger

• 
  @adamnachete30376 ай бұрын

  thank you 🙂🙂