giải hệ phg trình bằng phương pháp đặt ẩn phụ: 2/x+2y +1/y+2x =3 4/x+2y -3/y+2x =1

Đáp án: $\left( {x;y} \right) = \left( {\dfrac{1}{3};\dfrac{1}{3}} \right)$

Giải thích các bước giải:

$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x + 2y \ne 0\\
y + 2x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne  - 2y\\
y \ne  - 2x
\end{array} \right.\\
Dat:\dfrac{1}{{x + 2y}} = a;\dfrac{1}{{y + 2x}} = b\\
\left\{ \begin{array}{l}
\dfrac{2}{{x + 2y}} + \dfrac{1}{{y + 2x}} = 3\\
\dfrac{4}{{x + 2y}} - \dfrac{3}{{y + 2x}} = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2a + b = 3\\
4a - 3b = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
4a + 2b = 6\\
4a - 3b = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
5b = 5\\
2a + b = 3
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
b = 1\\
2a = 3 - b = 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
b = 1\\
a = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{{x + 2y}} = 1\\
\dfrac{1}{{y + 2x}} = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 2y = 1\\
y + 2x = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 2y = 1\\
2x + y = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x + 4y = 2\\
2x + y = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
3y = 1\\
x + 2y = 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = \dfrac{1}{3}\\
x = 1 - 2y = 1 - 2.\dfrac{1}{3} = \dfrac{1}{3}
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {\dfrac{1}{3};\dfrac{1}{3}} \right)
\end{array}$