x^3+x=0 2x(x-9)+3(x-9) =0 6x^2-3x=0 5x^3(7x+1)-10x^2(7x+1)=0

x^3+x=0
2x(x-9)+3(x-9) =0
6x^2-3x=0
5x^3(7x+1)-10x^2(7x+1)=0

0 bình luận về “x^3+x=0 2x(x-9)+3(x-9) =0 6x^2-3x=0 5x^3(7x+1)-10x^2(7x+1)=0”

  1. Đáp án + Giải thích các bước giải:

    `a,x^3+x=0`

    `=> x(x^2+1)=0`

    \(⇒\left[ \begin{array}{l}x=0\\x^2+1=0\end{array} \right.\)

    \(⇒\left[ \begin{array}{l}x=0\quad(Tm)\\x^2=-1\quad(Ktm)\end{array} \right.\)

    Vậy `S={0}`

    `b,2x(x-9)+3(x-9)=0`

    `=> (x-9)(2x+3)=0`

    \(⇒\left[ \begin{array}{l}x-9=0\\2x+3=0\end{array} \right.\)

    \(⇒\left[ \begin{array}{l}x=9\\2x=-3\end{array} \right.\)

    \(⇒\left[ \begin{array}{l}x=9\\x=-\dfrac{3}2\end{array} \right.\)

    Vậy `S={9;-3/2}`

    `c,6x^2-3x=0`

    `=>3x(2x-1)=0` 

    \(⇒\left[ \begin{array}{l}3x=0\\2x-1=0\end{array} \right.\)

    \(⇒\left[ \begin{array}{l}x=0\\2x=1\end{array} \right.\)

    \(⇒\left[ \begin{array}{l}x=0\\x=\dfrac{1}2\end{array} \right.\)

    Vậy `S={0;1/2}`

    `d,5x^3(7x+1)-10x^2(7x+1)=0`

    `=>(7x+1)(5x^3-10x^2)=0`

    `=>(7x+1)(x-2)5x^2=0`

    \(⇒\left[ \begin{array}{l}7x+1=0\\x-2=0\\5x^2=0\end{array} \right.\)

    \(⇒\left[ \begin{array}{l}7x=-1\\x=2\\x^2=0\end{array} \right.\)

    \(⇒\left[ \begin{array}{l}x=-\dfrac{1}7\\x=2\\x=0\end{array} \right.\)

    Vậy `S={-1/7;2;0}`

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  2. `x^3+x=0`

    `⇒x(x^2+1)=0`

    `⇒` \(\left[ \begin{array}{l}x=0\\x^2+1=0\end{array} \right.\)

    `⇒` \(\left[ \begin{array}{l}x=0\\x^2=-1\text{ (loại; do x² ≥ 0)}\end{array} \right.\)

    Vậy `x=0`.

    `2x(x-9)+3(x-9) =0`

    `⇒(2x+3)(x-9)=0`

    `⇒` \(\left[ \begin{array}{l}2x+3=0\\x-9=0\end{array} \right.\)

    $⇒ \left[ \begin{array}{l}x=9\\x=-\dfrac{3}2\end{array} \right.$

    Vậy `x=-3/2` hoặc `x=9`.

    `6x^2-3x=0`

    `⇒3x(2x-1)=0`

    `⇒` $\left[ \begin{array}{l}3x=0\\2x-1=0\end{array} \right.$

    $⇒\left[ \begin{array}{l}x=0\\x=\dfrac{1}2\end{array} \right.$

    Vậy `x=0` hoặc `x=1/2`.

    `5x^3(7x+1)-10x^2(7x+1)=0`

    `⇒(7x+1)(5x^3-10x^2)=0`

    `⇒(7x+1)(x-2)5x^2=0`

    $⇒\left[ \begin{array}{l}7x+1=0\\x-2=0\\5x^2=0\end{array} \right.$

    $⇒\left[ \begin{array}{l}7x=-1\\x=2\\x^2=0\end{array} \right.$

    $⇒\left[ \begin{array}{l}x=-\dfrac{1}7\\x=2\\x=0\end{array} \right.$

    Vậy `x∈{-1/7;2;0}`.

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