Toán Lớp 9: Tìm x,y biết:
$\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{6}{8}$
$\frac{4}{x}$ + $\frac{6}{y}$ = $\frac{2}{5}$
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Toán Lớp 9: Tìm x,y biết:
$\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{6}{8}$
$\frac{4}{x}$ + $\frac{6}{y}$ = $\frac{2}{5}$
Comments ( 1 )
\left\{ \begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{6}{8} = \dfrac{3}{4}\\
\dfrac{4}{x} + \dfrac{6}{y} = \dfrac{2}{5}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{4}{x} + \dfrac{4}{y} = 3\\
\dfrac{4}{x} + \dfrac{6}{y} = \dfrac{2}{5}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{4}{x} + \dfrac{6}{y} – \dfrac{4}{x} – \dfrac{4}{y} = \dfrac{2}{5} – 3\\
\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{3}{4}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{2}{y} = \dfrac{{ – 13}}{5}\\
\dfrac{1}{x} = \dfrac{3}{4} – \dfrac{1}{y}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{y} = \dfrac{{ – 13}}{{10}}\\
\dfrac{1}{x} = \dfrac{{41}}{{20}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{20}}{{41}}\\
y = – \dfrac{{10}}{{13}}
\end{array} \right.\\
Vậy\,x = \dfrac{{20}}{{41}};y = – \dfrac{{10}}{{13}}
\end{array}$