giúp mik bài 3b với ạ!!!! cần gấp ạ :<

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giúp mik bài 3b với ạ!!!!
cần gấp ạ :<
giup-mik-bai-3b-voi-a-can-gap-a

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Edana Edana 1 year 2020-11-16T01:30:50+00:00 1 Answers 56 views 0

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    2020-11-16T01:32:26+00:00

    Đáp án:

    $\begin{array}{l}
    b)Dkxd:\left\{ \begin{array}{l}
    4x + 1 \ge 0\\
    3x – 2 \ge 0
    \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
    x \ge  – \dfrac{1}{4}\\
    x \ge \dfrac{2}{3}
    \end{array} \right. \Rightarrow x \ge \dfrac{2}{3}\\
    \sqrt {4x + 1}  – \sqrt {3x – 2}  = \dfrac{{x + 3}}{5}\\
     \Leftrightarrow \dfrac{{4x + 1 – \left( {3x – 2} \right)}}{{\sqrt {4x + 1}  + \sqrt {3x – 2} }} – \dfrac{{x + 3}}{5} = 0\\
     \Leftrightarrow \dfrac{{x + 3}}{{\sqrt {4x + 1}  + \sqrt {3x – 2} }} – \dfrac{{x + 3}}{5} = 0\\
     \Leftrightarrow \left( {x + 3} \right)\left( {\dfrac{1}{{\sqrt {4x + 1}  + \sqrt {3x – 2} }} – \dfrac{1}{5}} \right) = 0\\
     \Rightarrow \left[ \begin{array}{l}
    x + 3 = 0\\
    \sqrt {4x + 1}  + \sqrt {3x – 2}  = 5
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x =  – 3\left( {ktm} \right)\\
    4x + 1 + 2\sqrt {4x + 1} .\sqrt {3x – 2}  + 3x – 2 = 25
    \end{array} \right.\\
     \Rightarrow 7x – 1 + 2\sqrt {12{x^2} – 5x – 2}  = 25\\
     \Rightarrow 2\sqrt {12{x^2} + 5x – 2}  = 26 – 7x\\
    \left( {dk:\dfrac{2}{3} \le x \le \dfrac{{26}}{7}} \right)\\
     \Rightarrow 4\left( {12{x^2} + 5x – 2} \right) = {\left( {26 – 7x} \right)^2}\\
     \Rightarrow 48{x^2} + 20x – 8 = 49{x^2} – 364x + 676\\
     \Rightarrow {x^2} – 384x + 684 = 0\\
     \Rightarrow \left[ \begin{array}{l}
    x = 1,789\left( {tm} \right)\\
    x = 382,21\left( {ktm} \right)
    \end{array} \right.\\
    Vay\,x = 1,789
    \end{array}$

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