Đáp án:
$\left( {x,y} \right) = \left( {\dfrac{{20}}{3},5} \right)$
Giải thích các bước giải:
Dựa vào đề bài ta có:
$\begin{array}{l}
\dfrac{{x + y}}{{\dfrac{1}{{20}}}} = \dfrac{{x - y}}{{\dfrac{1}{{140}}}} = \dfrac{{xy}}{{\dfrac{1}{7}}}\\
\Leftrightarrow \dfrac{{x + y}}{7} = \dfrac{{x - y}}{1} = \dfrac{{xy}}{{20}}\\
\Rightarrow \dfrac{{x + y}}{7} = \dfrac{{x - y}}{1} = \dfrac{{xy}}{{20}} = \dfrac{{x + y + x - y}}{{7 + 1}} = \dfrac{x}{4}\\
\Rightarrow \dfrac{{xy}}{{20}} = \dfrac{x}{4}\\
\Leftrightarrow xy - 5x = 0\\
\Leftrightarrow x\left( {y - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
y = 5
\end{array} \right.
\end{array}$
Ta xét trường hợp:
$\begin{array}{l}
+ )x = 0\\
\dfrac{{x + y}}{7} = \dfrac{{x - y}}{1}\\
\Leftrightarrow \dfrac{y}{7} = \dfrac{{ - y}}{1}\\
\Leftrightarrow y = 0\left( {l,do:x > y} \right)\\
+ )y = 5\\
\dfrac{{x + y}}{7} = \dfrac{{x - y}}{1}\\
\Leftrightarrow \dfrac{{x + 5}}{7} = \dfrac{{x - 5}}{1}\\
\Leftrightarrow x + 5 - 7\left( {x - 5} \right) = 0\\
\Leftrightarrow 40 - 6x = 0\\
\Leftrightarrow x = \dfrac{{20}}{3}\left( {tm} \right)
\end{array}$
Vậy $\left( {x,y} \right) = \left( {\dfrac{{20}}{3},5} \right)$
