x + 1/2x + 1 = 0,5x + 2/x+3 x-1/2005 = 3-y/2006 và x – y = 4009 Tìm x,y

Đáp án:

$\begin{array}{l}
\dfrac{{x + 1}}{{2x + 1}} = \dfrac{{0,5x + 2}}{{x + 3}} = \dfrac{{2.\left( {0,5x + 2} \right)}}{{2.\left( {x + 3} \right)}}\\
 = \dfrac{{x + 4}}{{2x + 6}} = \dfrac{{x + 1 – \left( {x + 4} \right)}}{{2x + 1 – \left( {2x + 6} \right)}} = \dfrac{{ – 3}}{{ – 5}} = \dfrac{3}{5}\\
 \Rightarrow \dfrac{{x + 1}}{{2x + 1}} = \dfrac{3}{5}\\
 \Rightarrow 5\left( {x + 1} \right) = 3.\left( {2x + 1} \right)\\
 \Rightarrow 5x + 5 = 6x + 3\\
 \Rightarrow 6x – 5x = 5 – 3\\
 \Rightarrow x = 2\\
Vay\,x = 2\\
\dfrac{{x – 1}}{{2005}} = \dfrac{{3 – y}}{{2006}}\\
 = \dfrac{{x – 1 + 3 – y}}{{2005 + 2006}} = \dfrac{{x – y + 2}}{{4011}}\\
 = \dfrac{{4009 + 2}}{{4011}} = 1\left( {do:x – y = 4009} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
x – 1 = 2005\\
3 – y = 2006
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 2006\\
y =  – 2003
\end{array} \right.\\
Vay\,x = 2006;y =  – 2003
\end{array}$