Giải hệ phương trình: \begin{cases}\frac{144}{x+y} + \frac{144}{x-y} = 21\\\frac{144}{x+y} + \frac{108}{x-y} = \frac{36}{y}\end{cases}

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1. Đáp án: $\left( x;y \right)=\left( 14;2 \right)$

   Giải thích:

    

   $\begin{cases}\dfrac{144}{x+y}+\dfrac{144}{x-y}=21\,\,\,\,\,\left(1\right)\\\dfrac{144}{x+y}+\dfrac{108}{x-y}=\dfrac{36}{y}\,\,\,\,\,\left(2\right)\end{cases}$

    

   Điều kiện:

   $\begin{cases}x+y\ne 0\\x-y\ne 0\\y\ne 0\end{cases}$

    

   Lấy $\left( 1 \right)-\left( 2 \right)$, ta được:

   $\,\,\,\,\,\,\,\dfrac{36}{x-y}=21+\dfrac{36}{y}$

    

   $\Leftrightarrow \dfrac{36}{x-y}=\dfrac{21y+36}{y}$

    

   $\Leftrightarrow 36y=\left( 21y+36 \right)\left( x-y \right)$

    

   $\Leftrightarrow 21xy-36x=21{{y}^{2}}$

   $\Leftrightarrow 7xy-12x=7{{y}^{2}}$

   $\Leftrightarrow x\left( 7y-12 \right)=7{{y}^{2}}$

   $\Leftrightarrow x=\dfrac{7{{y}^{2}}}{7y-12}$

    

   Phương trình $\left( 1 \right)$

   $\,\,\,\,\,\,\,144\left( \dfrac{1}{x+y}-\dfrac{1}{x-y} \right)=21$

    

   $\Leftrightarrow \dfrac{1}{x+y}-\dfrac{1}{x-y}=\dfrac{7}{48}$

    

   $\Leftrightarrow \dfrac{2x}{{{x}^{2}}-{{y}^{2}}}=\dfrac{7}{48}$

    

   $\Leftrightarrow 96x-7{{x}^{2}}+7{{y}^{2}}=0$

   $\Leftrightarrow 96\left( \dfrac{7{{y}^{2}}}{7y-12} \right)\,-\,7\,{{\left( \dfrac{7{{y}^{2}}}{7y-12} \right)}^{2}}+7{{y}^{2}}=0$

    

   $\Leftrightarrow 7{{y}^{2}}\left[ \dfrac{96}{7y-12}\,-\,\dfrac{49{{y}^{2}}}{{{\left( 7y-12 \right)}^{2}}}\,+\,1 \right]=0$

    

   $\Leftrightarrow \dfrac{96}{7y-12}\,-\,\dfrac{49{{y}^{2}}}{{{\left( 7y-12 \right)}^{2}}}+1=0$

    

   $\Leftrightarrow \dfrac{96\left( 7y-12 \right)}{{{\left( 7y-12 \right)}^{2}}}\,-\,\dfrac{49{{y}^{2}}}{{{\left( 7y-12 \right)}^{2}}}\,+\,\dfrac{{{\left( 7y-12 \right)}^{2}}}{{{\left( 7y-12 \right)}^{2}}}=0$

    

   $\Leftrightarrow 96\left( 7y-12 \right)-49{{y}^{2}}+{{\left( 7y-12 \right)}^{2}}=0$

   $\Leftrightarrow 504y=1008$

   $\Leftrightarrow y=2$

    

   $\to x=\dfrac{7{{y}^{2}}}{7y-12}=\dfrac{7\,.\,{{2}^{2}}}{7.2\,-\,12}=14$

    

   Thử lại ta thấy $\left( x;y \right)=\left( 14;2 \right)$ thỏa hệ phương trình

    

   Vậy $\left( x;y \right)=\left( 14;2 \right)$

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