Cho (3x -2y)/2=(2z-4x)/3 =(4y-3z)/2 tính B=(xy+yz+zx)/x^2+y^2+z^2

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1. Đáp án:

    $B= \dfrac{{26}}{{25}}$

   Giải thích các bước giải:

   \(\begin{array}{l}
   \dfrac{{3x – 2y}}{2} = \dfrac{{2z – 4x}}{3} = \dfrac{{4y – 3z}}{2}\\
    \Rightarrow \dfrac{{2z – 4x}}{3} = \dfrac{{6x – 4y}}{4} = \dfrac{{4y – 3z}}{2}\\
    = \dfrac{{6x – 4y + 4y – 3z}}{6} = \dfrac{{2z – 4x}}{3}\\
    \Rightarrow \dfrac{{2x – z}}{2} = \dfrac{{2z – 4x}}{3}\\
    \Rightarrow 6x – 3z = 4z – 8x \Rightarrow z = 2x\\
    + )\,\dfrac{{3x – 2y}}{2} = \dfrac{{6z – 12x}}{9} = \dfrac{{8y – 6z}}{4}\\
    = \dfrac{{6z – 12x + 8y – 6z}}{{13}}\\
    \Rightarrow \dfrac{{3x – 2y}}{2} = \dfrac{{ – 12x + 8y}}{{13}}\\
    \Rightarrow 39x – 26y =  – 24x + 16y\\
    \Leftrightarrow 63x = 42y \Rightarrow y = \dfrac{3}{2}x\\
    \Rightarrow B = \dfrac{{x.\dfrac{3}{2}x + \dfrac{3}{2}x.2x + 2x.x}}{{{x^2} + {{\left( {\dfrac{3}{2}x} \right)}^2} + {{\left( {2x} \right)}^2}}}\\
    = \dfrac{{\dfrac{{13}}{2}{x^2}}}{{\dfrac{{25}}{4}{x^2}}} = \dfrac{{26}}{{25}}
   \end{array}\)

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