Giúp em Bài 4 và bài 5 ạ

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Giúp em Bài 4 và bài 5 ạ
giup-em-bai-4-va-bai-5-a

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Latifah 4 years 2020-10-29T07:07:09+00:00 1 Answers 69 views 0

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  1. Sigrid
    0
    2020-10-29T07:09:06+00:00
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    Giải thích các bước giải:

    Ta có:

    \(\begin{array}{l}
    4,\\
    x + \dfrac{1}{x} = 3\\
    a,\\
    A = {x^2} + \dfrac{1}{{{x^2}}} = \left( {{x^2} + 2.x.\dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right) – 2.x.\dfrac{1}{x}\\
     = {\left( {x + \dfrac{1}{x}} \right)^2} – 2\\
     = {3^2} – 2\\
     = 7\\
    b,\\
    B = {x^3} + \dfrac{1}{{{x^3}}}\\
     = \left( {{x^3} + 3.{x^2}.\dfrac{1}{x} + 3.x.\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}} \right) – \left( {3.{x^2}.\dfrac{1}{x} + 3.x.\dfrac{1}{{{x^2}}}} \right)\\
     = {\left( {x + \dfrac{1}{x}} \right)^3} – 3.x.\dfrac{1}{x}.\left( {x + \dfrac{1}{x}} \right)\\
     = {3^3} – 3.1.3\\
     = 18\\
    5,\\
    {x^2} + {y^2} – 4x – 6y + 13 = 0\\
     \Leftrightarrow \left( {{x^2} – 4x + 4} \right) + \left( {{y^2} – 6y + 9} \right) = 0\\
     \Leftrightarrow {\left( {x – 2} \right)^2} + {\left( {y – 3} \right)^2} = 0\\
     \Leftrightarrow \left\{ \begin{array}{l}
    {\left( {x – 2} \right)^2} = 0\\
    {\left( {y – 3} \right)^2} = 0
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x = 2\\
    y = 3
    \end{array} \right.\\
    S = {\left( {x – y} \right)^{2020}} = {\left( {2 – 3} \right)^{2020}} = {\left( { – 1} \right)^{2020}} = 1
    \end{array}\)

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