Toán Giải hệ pt điều kiện loại 2 {x+xy-y=1 x^2 +y^2 – 4(x-y) = -6 24/08/2021 By Arya Giải hệ pt điều kiện loại 2 {x+xy-y=1 x^2 +y^2 – 4(x-y) = -6
Đáp án: \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x = 3\\y = – 1\end{array} \right.\\\left\{ \begin{array}{l}x = 1\\y = – 3\end{array} \right.\\\left\{ \begin{array}{l}x = 1\\y = – 1\end{array} \right.\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}\left\{ \begin{array}{l}x + xy – y = 1\\{x^2} + {y^2} – 4(x – y) = – 6\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}xy + (x – y) = 1\\{(x – y)^2} – 4(x – y) + 2xy = – 6\end{array} \right.\\x – y = a,xy = b\\ \Rightarrow \left\{ \begin{array}{l}b + a = 1\\{a^2} – 4a + 2b = – 6\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 1 – a\\{a^2} – 4a + 2(1 – a) = – 6\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 1 – a\\\left[ \begin{array}{l}a = 4\\a = 2\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}a = 4,b = – 3\\a = 2,b = – 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x – y = 4\\xy = – 3\end{array} \right.\\\left\{ \begin{array}{l}x – y = 2\\xy = – 1\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 3\\y = – 1\end{array} \right.\\\left\{ \begin{array}{l}x = 1\\y = – 3\end{array} \right.\\\left\{ \begin{array}{l}x = 1\\y = – 1\end{array} \right.\end{array} \right.\end{array}\) Trả lời
Đáp án:
Giải thích các bước giải:
Đáp án:
\(\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 3\\
y = – 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y = – 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y = – 1
\end{array} \right.
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + xy – y = 1\\
{x^2} + {y^2} – 4(x – y) = – 6
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
xy + (x – y) = 1\\
{(x – y)^2} – 4(x – y) + 2xy = – 6
\end{array} \right.\\
x – y = a,xy = b\\
\Rightarrow \left\{ \begin{array}{l}
b + a = 1\\
{a^2} – 4a + 2b = – 6
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 1 – a\\
{a^2} – 4a + 2(1 – a) = – 6
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 1 – a\\
\left[ \begin{array}{l}
a = 4\\
a = 2
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a = 4,b = – 3\\
a = 2,b = – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – y = 4\\
xy = – 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x – y = 2\\
xy = – 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 3\\
y = – 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y = – 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y = – 1
\end{array} \right.
\end{array} \right.
\end{array}\)