giai he phuong trinh x^2 +xy-2y^2=0 va 3x+2y=5xy

Đáp án:

\[\left( {x;y} \right) = \left\{ {\left( { – \frac{4}{5};\frac{2}{5}} \right);\left( {0;0} \right);\left( {1;1} \right)} \right\}\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + xy – 2{y^2} = 0\,\,\,\,\,\,\,\,\left( 1 \right)\\
3x + 2y = 5xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow {x^2} + xy – 2{y^2} = 0\\
 \Leftrightarrow \left( {{x^2} – xy} \right) + \left( {2xy – 2{y^2}} \right) = 0\\
 \Leftrightarrow x\left( {x – y} \right) + 2y\left( {x – y} \right) = 0\\
 \Leftrightarrow \left( {x – y} \right)\left( {x + 2y} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x – y = 0\\
x + 2y = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = y\\
x =  – 2y
\end{array} \right.\\
TH1:\,\,\,\,\,x = y\\
\left( 2 \right) \Leftrightarrow 3x + 2x = 5x.x\\
 \Leftrightarrow 5{x^2} – 5x = 0\\
 \Leftrightarrow {x^2} – x = 0\\
 \Leftrightarrow x\left( {x – 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = y = 0\\
x = y = 1
\end{array} \right.\\
TH2:\,\,\,\,x =  – 2y\\
3.\left( { – 2y} \right) + 2y = 5.\left( { – 2y} \right).y\\
 \Leftrightarrow  – 6y + 2y =  – 10{y^2}\\
 \Leftrightarrow 10{y^2} – 4y = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
y = 0 \Rightarrow x = 0\\
y = \frac{2}{5} \Rightarrow x =  – \frac{4}{5}
\end{array} \right.\\
 \Rightarrow \left( {x;y} \right) = \left\{ {\left( { – \frac{4}{5};\frac{2}{5}} \right);\left( {0;0} \right);\left( {1;1} \right)} \right\}
\end{array}\)