Can You Solve This Radical Algebra Challenge?

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  • Опубликовано: 27 янв 2025

Комментарии • 8

  • @virat-k9g
    @virat-k9g 13 дней назад +1

    Can anyone tell me the value of the given expression. 😢

  • @jakesboty9936
    @jakesboty9936 14 дней назад +1

    a=5,b=2

  • @潘博宇-k4l
    @潘博宇-k4l 13 дней назад

    a=10, b=4.

  • @RashmiRay-c1y
    @RashmiRay-c1y 13 дней назад

    Let 2^1/3=u. So, u^3=2. x=u^2+u+1= (u^3-1)/(u-1) = 1/(u-1). So, 1/x=u-1. Therefore, 1/x^4+3/x^3+1/x^2+3= u^4-u^3-2u^2+3u+2 = u^4-2u^2+3u= 5u-2u^2. Again, 1/u^2=u/2 and 1/u= u^2/2. So, 5u-2u^2=(a/2)u -(b/2)u^2. So, a=10, b=4.

  • @Fjfurufjdfjd
    @Fjfurufjdfjd 14 дней назад

    α=10 β=4

  • @gregevgeni1864
    @gregevgeni1864 14 дней назад

    x=³√4+³√2+1 =>
    x•(³√2-1)=(³√2²+³√2+1²)(³√2-1)=
    ³√2³-1³=2-1=1 => 1/x =³√2-1 =>
    1/x⁴ +3/x³ +1/x²+3=
    (1/x)⁴+3(1/x)³+(1/x)²+3=
    (³√2-1)⁴+3(³√2-1)³+(³√2-1)²+3=
    5•³√2 -2•³√4 =
    (5• ³√2•³√4)/³√4-(2•³√4•³√2)/³√2 =5•³√8/³√4-2•³√8/³√2=
    =5•2/³√4-2•2/³√2=10/³√4-4/³√2
    => a =10 and b =4.

  • @vcvartak7111
    @vcvartak7111 13 дней назад

    Terrible manipulation, has to be more systematic way. It's impossible to envisaged this jugglery in exam