giai he phuong trinh x^2 +xy-2y^2=0 va 3x+2y=5xy 13/07/2021 Bởi Ayla 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: \(\begin{array}{l}\left\{ \begin{array}{l}{x^2} + xy – 2{y^2} = 0\,\,\,\,\,\,\,\,\left( 1 \right)\\3x + 2y = 5xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\end{array} \right.\\\left( 1 \right) \Leftrightarrow {x^2} + xy – 2{y^2} = 0\\ \Leftrightarrow \left( {{x^2} – xy} \right) + \left( {2xy – 2{y^2}} \right) = 0\\ \Leftrightarrow x\left( {x – y} \right) + 2y\left( {x – y} \right) = 0\\ \Leftrightarrow \left( {x – y} \right)\left( {x + 2y} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x – y = 0\\x + 2y = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = y\\x = – 2y\end{array} \right.\\TH1:\,\,\,\,\,x = y\\\left( 2 \right) \Leftrightarrow 3x + 2x = 5x.x\\ \Leftrightarrow 5{x^2} – 5x = 0\\ \Leftrightarrow {x^2} – x = 0\\ \Leftrightarrow x\left( {x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = y = 0\\x = y = 1\end{array} \right.\\TH2:\,\,\,\,x = – 2y\\3.\left( { – 2y} \right) + 2y = 5.\left( { – 2y} \right).y\\ \Leftrightarrow – 6y + 2y = – 10{y^2}\\ \Leftrightarrow 10{y^2} – 4y = 0\\ \Leftrightarrow \left[ \begin{array}{l}y = 0 \Rightarrow x = 0\\y = \frac{2}{5} \Rightarrow x = – \frac{4}{5}\end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left\{ {\left( { – \frac{4}{5};\frac{2}{5}} \right);\left( {0;0} \right);\left( {1;1} \right)} \right\}\end{array}\) Bình luận
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:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + xy – 2{y^2} = 0\,\,\,\,\,\,\,\,\left( 1 \right)\\
3x + 2y = 5xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow {x^2} + xy – 2{y^2} = 0\\
\Leftrightarrow \left( {{x^2} – xy} \right) + \left( {2xy – 2{y^2}} \right) = 0\\
\Leftrightarrow x\left( {x – y} \right) + 2y\left( {x – y} \right) = 0\\
\Leftrightarrow \left( {x – y} \right)\left( {x + 2y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – y = 0\\
x + 2y = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = y\\
x = – 2y
\end{array} \right.\\
TH1:\,\,\,\,\,x = y\\
\left( 2 \right) \Leftrightarrow 3x + 2x = 5x.x\\
\Leftrightarrow 5{x^2} – 5x = 0\\
\Leftrightarrow {x^2} – x = 0\\
\Leftrightarrow x\left( {x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = y = 0\\
x = y = 1
\end{array} \right.\\
TH2:\,\,\,\,x = – 2y\\
3.\left( { – 2y} \right) + 2y = 5.\left( { – 2y} \right).y\\
\Leftrightarrow – 6y + 2y = – 10{y^2}\\
\Leftrightarrow 10{y^2} – 4y = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y = 0 \Rightarrow x = 0\\
y = \frac{2}{5} \Rightarrow x = – \frac{4}{5}
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( { – \frac{4}{5};\frac{2}{5}} \right);\left( {0;0} \right);\left( {1;1} \right)} \right\}
\end{array}\)