-7x^2+4/x^3+1=5/x^2-x+1-1/x+1 x-2/x+2-3/x-2=2(x-11)/x^2-4 x+5/x-1-x+1/x-3=-8/x^2-4x+3

0 bình luận về “-7x^2+4/x^3+1=5/x^2-x+1-1/x+1 x-2/x+2-3/x-2=2(x-11)/x^2-4 x+5/x-1-x+1/x-3=-8/x^2-4x+3”

1. Đáp án:

    c) Phương trình vô nghiệm

   Giải thích các bước giải:

   \(\begin{array}{l}
   a)DK:x \ne  – 1\\
   \dfrac{{ – 7{x^2} + 4}}{{{x^3} + 1}} = \dfrac{5}{{{x^2} – x + 1}} – \dfrac{1}{{x + 1}}\\
    \to \dfrac{{ – 7{x^2} + 4}}{{{x^3} + 1}} = \dfrac{{5\left( {x + 1} \right) – {x^2} + x – 1}}{{{x^3} + 1}}\\
    \to  – 7{x^2} + 4 = 5x + 5 – {x^2} + x – 1\\
    \to 6{x^2} + 6x = 0\\
    \to 6x\left( {x + 1} \right) = 0\\
    \to \left[ \begin{array}{l}
   x = 0\left( {TM} \right)\\
   x =  – 1\left( l \right)
   \end{array} \right.\\
   b)DK:x \ne  \pm 2\\
   \dfrac{{x – 2}}{{x + 2}} – \dfrac{3}{{x – 2}} = \dfrac{{2x – 22}}{{{x^2} – 4}}\\
    \to \dfrac{{{{\left( {x – 2} \right)}^2} – 3\left( {x + 2} \right) – 2x + 22}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = 0\\
    \to \dfrac{{{x^2} – 4x + 4 – 3x – 6 – 2x + 22}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = 0\\
    \to {x^2} – 9x + 20 = 0\\
    \to {x^2} – 4x – 5x + 20 = 0\\
    \to x\left( {x – 4} \right) – 5\left( {x – 4} \right) = 0\\
    \to \left( {x – 4} \right)\left( {x – 5} \right) = 0\\
    \to \left[ \begin{array}{l}
   x = 4\\
   x = 5
   \end{array} \right.\\
   c)DK:x \ne \left\{ {1;3} \right\}\\
   \dfrac{{x + 5}}{{x – 1}} – \dfrac{{x + 1}}{{x – 3}} = \dfrac{{ – 8}}{{{x^2} – 4x + 3}}\\
    \to \dfrac{{\left( {x + 5} \right)\left( {x – 3} \right) – \left( {x + 1} \right)\left( {x – 1} \right) + 8}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
    \to \dfrac{{{x^2} + 2x – 15 – {x^2} + 1 + 8}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
    \to 2x – 6 = 0\\
    \to x = 3\left( l \right)
   \end{array}\)

   ⇒ Phương trình vô nghiệm

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