Harvard University Admission Interview Tricks.✍️🖋️📘💙

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  • Опубликовано: 31 дек 2024

Комментарии • 9

  • @mohammedkamal3365
    @mohammedkamal3365 3 дня назад +2

    شكرا : شرح واضح ومنطقي .

    • @superacademy247
      @superacademy247  3 дня назад +1

      I'm glad you found it helpful 🌲🎄🎉🥳🎁

  • @davidseed2939
    @davidseed2939 3 дня назад +1

    quicker and easier if you calculated y^6 first
    then set m=y^6/729
    on the way calculate y^2 and y^3 and check
    3y^2 +y^3
    = (12-6_/3) +(-10 +6_/3)
    =2 as required.
    y^6= (-10+6_/3)^2
    =208-120_/3
    m=y^6/729
    =208/729 -40/243

  • @اسماعیلخسروی-خ6ظ
    @اسماعیلخسروی-خ6ظ 3 дня назад +1

    ❤❤

  • @ichdu6710
    @ichdu6710 2 дня назад

    other way to solve the equation y³ + y² - 2/27 = 0
    1. Find a solution by inspection. I found one after trying three fractions
    2. the solution ist y = - 1/3
    3. factorize y³ + y² - 2/27 = ( y + 1/3 )( y² + ay - 2/9 )
    4. the factorizing can be found by long division --> a =+2/3
    5. solving ( y + 1/3 )( y² + 2y/3 - 2/9 ) =0
    6. solution 1: y = - 1/3 isn't a solution, while y must be positive
    7 solution 2: y =- (-1-root(3))/3 isn't a solution, while y must be positive
    8. solution 3: y -= (-1+root(3))/3 is a solution, while y is positive
    9. for y ^6 calculate {(-1+root(3))/3]² * {(-1+root(3))/3]² * {(-1+root(3))/3]²

  • @savery1967
    @savery1967 3 дня назад +1

    y can be negative!
    m must be positive but a negative y can still give a positive m; m=y^6

    • @superacademy247
      @superacademy247  3 дня назад

      Thanks for your nice question. A negative y is inconsistent to the domain of m. Therefore negative y MUST be disregarded.🥳🎉🎁🎄🌲

  • @ИринаСавостин
    @ИринаСавостин 3 дня назад

    10^х+100^х=1000^х
    10^х+10^2х=10^3х
    10^х*(1+10^х)=10^3
    Делим обе части равенства на 10^х, 10^x не равен 0 в любом случае
    1+10^х=10^2х
    10^2х-10^х-1=0
    Замена10^х=а
    а^2-а-1=0
    Ищем корни
    Д=(-1)^2-4*1*(-1)=1+4=5
    а=(1+5^(1/2))/2
    или а=(1-5^(1/2))/2
    Ищем х
    1 случай
    10^х=(1+5^(1/2))/2
    lg10^x=lg((1+5^(1/2))/2) lg10^x=x*lg10=x*1=x
    x=lg(1+5^(1/2)/2)
    lgb,b>o