Giúp em giải với ạ.pls

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Giúp em giải với ạ.pls
giup-em-giai-voi-a-pls

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Acacia 6 years 2020-11-21T15:45:48+00:00 2 Answers 59 views 0

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  1. Verity
    0
    2020-11-21T15:47:26+00:00
    Reply

    Đáp án:

    i. \(TXD:D = R\backslash \left\{ { – 1} \right\}\)

    Giải thích các bước giải:

    \(\begin{array}{l}
    a.DK:2{x^2} – x – 1 \ne 0\\
     \to x \ne \left\{ { – \dfrac{1}{2};1} \right\}\\
     \to TXD:D = R\backslash \left\{ { – \dfrac{1}{2};1} \right\}\\
    d.DK:{x^3} – x \ne 0\\
     \to x \ne \left\{ { – 1;0;1} \right\}\\
     \to TXD:D = R\backslash \left\{ { – 1;0;1} \right\}\\
    g.DK:2{x^2} – x + 2 \ne 0\left( {ld} \right)\\
    Do:2{x^2} – x + 2 > 0\forall x \in R\\
     \to TXD:D = R\\
    b.DK:{x^2} – x – 2 \ne 0\\
     \to x \ne \left\{ { – 1;2} \right\}\\
     \to TXD:D = R\backslash \left\{ { – 1;2} \right\}\\
    e.DK:{x^2} + 3x \ne 0\\
     \to x \ne \left\{ { – 3;0} \right\}\\
     \to TXD:D = R\backslash \left\{ { – 3;0} \right\}\\
    h.DK: – 6x + 3 \ne 0\\
     \to x \ne \dfrac{1}{2}\\
     \to TXD:D = R\backslash \dfrac{1}{2}\\
    c.DK:\left( {x + 4} \right)\left( {x – 1} \right) \ne 0\\
     \to x \ne \left\{ { – 4;1} \right\}\\
     \to TXD:D = R\backslash \left\{ { – 4;1} \right\}\\
    f.DK:{x^2} – 3x + 12 \ne 0\left( {ld} \right)\\
    Do:{x^2} – 3x + 12 > 0\forall x \in R\\
     \to TXD:D = R\\
    i.DK:3{x^2} + 6x + 3 \ne 0\\
     \to x \ne  – 1\\
     \to TXD:D = R\backslash \left\{ { – 1} \right\}
    \end{array}\)

  2. Doris
    0
    2020-11-21T15:47:39+00:00
    Reply

    `a) 2x² – x – 1 ne 0`

    `<=> 2x² – 2x + x – 1 ne 0`

    `<=> 2x(x – 1) + (x – 1) ne 0`

    `<=> (2x + 1)(x – 1) ne 0`

    `<=>` \(\left[ \begin{array}{l}x \ne \dfrac{-1}{2}\\x \ne 1\end{array} \right.\) 

    `=> D = RR` \ `{(-1)/2; 1}`

    `b) x³ – x ne 0`

    `<=> x(x² – 1) ne 0`

    `<=>` \(\left[ \begin{array}{l}x \ne 0\\x \ne ±1\end{array} \right.\) 

    `=> D = RR` \ `{0; ±1}`

    `c) 2x² – x + 2 \ne 0`

    `<=> 2(x² – 1/(2)x + 1) ne 0`

    `<=> 2(x – 1/2)² + 15/8 > 0` với `∀ x ∈ RR`

    `=> D = RR`

    `d) x² – x – 2 ne 0`

    `<=> x² + x – 2x – 2 ne 0`

    `<=> (x – 2)(x + 1) ne 0`

    `<=>` \(\left[ \begin{array}{l}x \ne 2\\x \ne -1\end{array} \right.\) 

    `=> D = RR` \ `{-1; 2}`

    e) x² + 3x ne 0`

    `<=> x(x + 3) ne 0`

    `<=>` \(\left[ \begin{array}{l}x \ne 0\\x \ne -3\end{array} \right.\) 

    `=> D = RR` \ `{0; -3}`

    `f) -6x + 3 ne 0`

    `<=> 2x – 1 ne 0`

    `<=> x ne 1/2`

    `=> D = RR` \ `{1/2}`

    `g) (x + 4)(x – 1) ne 0`

    `<=>` \(\left[ \begin{array}{l}x \ne -4\\x \ne 1\end{array} \right.\) 

    `=> D = RR` \ `{-4; 1}`

    `h) x² – 3x + 12 ne 0`

    `<=> (x – 3/2)² + 39/4 > 0` với `∀ x ∈ RR`

    `i) 3x² + 6x + 3 ne 0`

    `<=> x² + 2x + 1 ne 0`

    `<=> (x + 1)² ne 0`

    `=> x ne -1`

    `=> D = RR` \ `{-1}`

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