Đáp án:

Hệ phương trình vô nghiệm

Giải thích các bước giải:

\(\begin{array}{l}
DK:x \ne  \pm y;y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{{30}}{{x + y}} + \dfrac{{30}}{{x - y}} = \dfrac{5}{2}\\
\dfrac{{30}}{{x + y}} + \dfrac{{20}}{{x - y}} = \dfrac{{10}}{y}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{10}}{{x - y}} = \dfrac{5}{2} - \dfrac{{10}}{y}\\
\dfrac{{30}}{{x + y}} + \dfrac{{30}}{{x - y}} = \dfrac{5}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{10}}{{x - y}} = \dfrac{{5y - 20}}{{2y}}\\
\dfrac{{30}}{{x + y}} + \dfrac{{30}}{{x - y}} = \dfrac{5}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
20y = \left( {x - y} \right)\left( {5y - 20} \right)\\
\dfrac{{30}}{{x + y}} + \dfrac{{30}}{{x - y}} = \dfrac{5}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
4y = \left( {x - y} \right)\left( {y - 4} \right)\\
\dfrac{{30}}{{x + y}} + \dfrac{{30}}{{x - y}} = \dfrac{5}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{4y}}{{y - 4}} = x - y\\
\dfrac{{30}}{{x + y}} + \dfrac{{30}}{{x - y}} = \dfrac{5}{2}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{4y}}{{y - 4}} + y\\
\dfrac{{30}}{{\left( {\dfrac{{4y}}{{y - 4}} + y} \right) + y}} + \dfrac{{30}}{{\left( {\dfrac{{4y}}{{y - 4}} + y} \right) - y}} = \dfrac{5}{2}\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to \dfrac{{30}}{{\dfrac{{4y}}{{y - 4}} + 2y}} + \dfrac{{30}}{{\dfrac{{4y}}{{y - 4}}}} = \dfrac{5}{2}\\
 \to 30:\left( {\dfrac{{4y + 2{y^2} - 8y}}{{y - 4}}} \right) + 30:\left( {\dfrac{{4y}}{{y - 4}}} \right) = \dfrac{5}{2}\\
 \to 30:\left( {\dfrac{{2{y^2} - 4y}}{{y - 4}} + \dfrac{{4y}}{{y - 4}}} \right) = \dfrac{5}{2}\\
 \to \dfrac{{2{y^2}}}{{y - 4}} = 12\\
 \to 2{y^2} = 12y - 48\left( {DK:y \ne 4} \right)
\end{array}\)

⇒ Phương trình vô nghiệm

⇒ Hệ phương trình vô nghiệm

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