Tìm x biết: ($x^{3}$ -1):(x-1)-( $9x^{2}$ -1):(3x-1)=0

Tìm x biết:
($x^{3}$ -1):(x-1)-( $9x^{2}$ -1):(3x-1)=0

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  1. (x³ -1):(x-1)-(9x²-1):(3x-1)=0

    (=)(x -1)(x²+x+1):(x-1)-(3x-1)(3x+1):(3x-1)=0

    (=)x²+x+1-3x-1=0

    (=)x²-2x=0

    (=)x(x-2)=0

    (=)\(\left[ \begin{array}{l}x=0\\x-2=0\end{array} \right.\) (=)\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\) 

    S={2;0}

     

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  2. Đáp án:Tham khảo

     

    Giải thích các bước giải:

    $\text{ĐK}$:\(\left[ \begin{array}{l}:x-1\neq 0\\3x-1\neq 0\end{array} \right.\) 

    ⇔\(\left[ \begin{array}{l}x\neq 1\\x\neq \dfrac{1}{3}\end{array} \right.\) 
    $\text{Ta có:}$

    $(x³-1):(x-1)-(9x²-1):(3x-1)=0$

    $⇔(x-1)(x²+x+1):(x-1)-[(3x)²-1²]:(3x-1)=0$

    $⇔(x-1)(x²+x+1):(x-1)-(3x-1)(3x+1):(3x-1)=0$

    $⇔(x²+x+1)-(3x+1)=0$

    $⇔x²+x+1-3x-1=0$

    $⇔x²-2x=0$

    $⇔x(x-2)=0$

    ⇔\(\left[ \begin{array}{l}x=0\\x-2=0\end{array} \right.\) 

    ⇔\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\) 

    $\text{Vậy x∈(0;2)}$

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