cho x*2-5x+1=0. Tính x*2+1/x*2

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cho x*2-5x+1=0. Tính x*2+1/x*2

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Helga 6 years 2020-11-21T17:07:12+00:00 2 Answers 110 views 0

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  1. Nem
    0
    2020-11-21T17:08:58+00:00
    Reply

         x*2-5x+1  =0

    =>2x-5x +1+=0

    =>2x-5x        = 0 -1

    =>2x-5x        = -1

    => -3x            = -1

    =>     x            = 1/3

    Từ đó ta thay x = 1/3 để tính x*2+1/x*2 => Ta được bài toán mới: 

         1/3*2+1/(1/3*2)

    => 2/3+ 1/(2/3)

    => 2/3+3/2

    => 4/6+9/6 (MC:6)

    => 13/6

  2. Orla
    0
    2020-11-21T17:09:05+00:00
    Reply

    Đáp án:

    \({x_2} + \dfrac{1}{{{x_2}}} = 5\)

    Giải thích các bước giải:

    \(\begin{array}{l}
    Xét:\Delta  = 25 – 4.1 = 21 > 0\\
     \to \left[ \begin{array}{l}
    x = \dfrac{{5 + \sqrt {21} }}{2}\\
    x = \dfrac{{5 – \sqrt {21} }}{2}
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    {x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{5 + \sqrt {21} }}{2} + \dfrac{2}{{5 + \sqrt {21} }}\\
    {x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{5 – \sqrt {21} }}{2} + \dfrac{2}{{5 – \sqrt {21} }}
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    {x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{25 + 10\sqrt {21}  + 21 + 4}}{{2\left( {5 + \sqrt {21} } \right)}}\\
    {x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{25 – 10\sqrt {21}  + 21 + 4}}{{2\left( {5 – \sqrt {21} } \right)}}
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    {x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{50 + 10\sqrt {21} }}{{2\left( {5 + \sqrt {21} } \right)}}\\
    {x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{50 – 10\sqrt {21} }}{{2\left( {5 – \sqrt {21} } \right)}}
    \end{array} \right.\\
     \to \left[ \begin{array}{l}
    {x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{10\left( {5 + \sqrt {21} } \right)}}{{2\left( {5 + \sqrt {21} } \right)}} = 5\\
    {x_2} + \dfrac{1}{{{x_2}}} = \dfrac{{10\left( {5 – \sqrt {21} } \right)}}{{2\left( {5 – \sqrt {21} } \right)}} = 5
    \end{array} \right.\\
     \to {x_2} + \dfrac{1}{{{x_2}}} = 5
    \end{array}\)

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )