Home/ Intrebari/Q 69587
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

Agent-Bond
Pe: 2010-02-26T08:42:53+02:00 In: In:

Radical ordin 3

Sa se rezolve ecuatia:
rad ord 3(x) + rad ord 3 (2x-3) = rad ord 3 (12(x-1))

Similare

2 raspunsuri

  1. Alin

    Ecuatia arata cam asa :

    \sqrt[3]{x}+\sqrt[3]{2x-3}=\sqrt[3]{12(x-1)}

    M-am gandit sa notez a=x-1. Atunci ecuatia devine :

    \sqrt[3]{a+1}+\sqrt[3]{2a-1}=\sqrt[3]{12a}

    Ridicam la puterea a 3-a si obtinem :

    a+1+2a-1+3\sqrt[3]{(a+1)(2a-1)}(\sqrt[3]{a+1}+\sqrt[3]{2a-1})=12a \Longleftrightarrow 3\sqrt[3]{(a+1)(2a-1)}(\sqrt[3]{a+1}+\sqrt[3]{2a-1})=9a \Longleftrightarrow \sqrt[3]{(a+1)(2a-1)}(\sqrt[3]{a+1}+\sqrt[3]{2a-1})=3a (1)

    Dar \sqrt[3]{a+1}+\sqrt[3]{2a-1}=\sqrt[3]{12a} si inlocuind in (1) obtinem :

    \sqrt[3]{(a+1)(2a-1)12a}=3a

    Ridicam din nou la puterea a 3-a si obtinem :

    (a+1)(2a-1)12a=27a^3 \Longleftrightarrow (2a^2+a-1)12a=27a^3 \Longleftrightarrow 24a^3+12a^2-12a=27a^3 \Longleftrightarrow a^3=4a^2-4a .

    Daca a\ne 0 impartim prin a si obtinem :

    a^2=4a-4 \Longleftrightarrow a^2-4a+4=0 \Longleftrightarrow (a-2)^2=0 \Longleftrightarrow a=2 \Longleftrightarrow x=3 , care verifica.

    Se observa usor ca a=0 \Longleftrightarrow x=1 este solutie.

    Deci avem : S=\{ 1,3 \}.

  2. Shadow23

    Buna! Ajutati-ma, va rog. Am de rezolvat o problema dar m-am cam impotmolit.
    Sa se compare perechile de numere:
    2 radical de ordin 3 din 1 supra 54 si radical de ordin 3 din 8 supra 375 .

Raspunde

Raspunde