Giải hệ PT {x² + y² – x + y = 2 {xy + x – y = -1 Giúp với 05/12/2021 Bởi Valerie Giải hệ PT {x² + y² – x + y = 2 {xy + x – y = -1 Giúp với
Đáp án: \(\left[ \begin{array}{l}y = 0;x = – 1\\y = 1;x = 0\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}\left\{ \begin{array}{l}{x^2} + {y^2} – \left( {x – y} \right) = 2\\xy + x – y = – 1\end{array} \right.\\ \to \left\{ \begin{array}{l}{x^2} + {y^2} – \left( {x – y} \right) = 2\\x – y = – 1 – xy\end{array} \right.\\ \to \left\{ \begin{array}{l}{x^2} + {y^2} – \left( { – 1 – xy} \right) = 2\\x – y = – 1 – xy\end{array} \right.\\ \to \left\{ \begin{array}{l}{x^2} + {y^2} + xy = 1\\x – y = – 1 – xy\end{array} \right.\\ \to \left\{ \begin{array}{l}{x^2} + {y^2} – 2xy + 3xy = 1\\x – y = – 1 – xy\end{array} \right.\\ \to \left\{ \begin{array}{l}{\left( {x – y} \right)^2} + 3xy = 1\\x – y = – 1 – xy\end{array} \right.\\ \to \left\{ \begin{array}{l}{\left( { – 1 – xy} \right)^2} + 3xy = 1\\x – y = – 1 – xy\end{array} \right.\\ \to \left\{ \begin{array}{l}1 + 2xy + {\left( {xy} \right)^2} + 3xy = 1\\x – y = – 1 – xy\end{array} \right.\\ \to \left\{ \begin{array}{l}{\left( {xy} \right)^2} + 5xy = 0\\x – y = – 1 – xy\end{array} \right.\\ \to \left[ \begin{array}{l}xy = 0\\xy = – 5\end{array} \right. \to \left[ \begin{array}{l}x – y = – 1\\x – y = 4\end{array} \right.\\ \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x = – 1 + y\\\left( { – 1 + y} \right).y = 0\end{array} \right.\\\left\{ \begin{array}{l}x = y + 4\\\left( {y + 4} \right).y = – 5\left( {vô nghiệm} \right)\end{array} \right.\end{array} \right. \to \left\{ \begin{array}{l}x = – 1 + y\\\left[ \begin{array}{l}y = 0\\y = 1\end{array} \right.\end{array} \right.\\ \to \left[ \begin{array}{l}y = 0;x = – 1\\y = 1;x = 0\end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
y = 0;x = – 1\\
y = 1;x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + {y^2} – \left( {x – y} \right) = 2\\
xy + x – y = – 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} + {y^2} – \left( {x – y} \right) = 2\\
x – y = – 1 – xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} + {y^2} – \left( { – 1 – xy} \right) = 2\\
x – y = – 1 – xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} + {y^2} + xy = 1\\
x – y = – 1 – xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} + {y^2} – 2xy + 3xy = 1\\
x – y = – 1 – xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{\left( {x – y} \right)^2} + 3xy = 1\\
x – y = – 1 – xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{\left( { – 1 – xy} \right)^2} + 3xy = 1\\
x – y = – 1 – xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
1 + 2xy + {\left( {xy} \right)^2} + 3xy = 1\\
x – y = – 1 – xy
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{\left( {xy} \right)^2} + 5xy = 0\\
x – y = – 1 – xy
\end{array} \right.\\
\to \left[ \begin{array}{l}
xy = 0\\
xy = – 5
\end{array} \right. \to \left[ \begin{array}{l}
x – y = – 1\\
x – y = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = – 1 + y\\
\left( { – 1 + y} \right).y = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x = y + 4\\
\left( {y + 4} \right).y = – 5\left( {vô nghiệm} \right)
\end{array} \right.
\end{array} \right. \to \left\{ \begin{array}{l}
x = – 1 + y\\
\left[ \begin{array}{l}
y = 0\\
y = 1
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 0;x = – 1\\
y = 1;x = 0
\end{array} \right.
\end{array}\)