tìm x,y biết +) x+y=-1 và x-657y=-1983 +) x+y=1 và x-657y=1983 +) x+y=-1983 và x-657y=-1 +) x+y=1983 và x-657y=1 06/10/2021 Bởi Samantha tìm x,y biết +) x+y=-1 và x-657y=-1983 +) x+y=1 và x-657y=1983 +) x+y=-1983 và x-657y=-1 +) x+y=1983 và x-657y=1
Đáp án: a) \(\left\{ \begin{array}{l}y = \dfrac{{991}}{{329}}\\x = – \dfrac{{1320}}{{329}}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a)\left\{ \begin{array}{l}x + y = – 1\\x – 657y = – 1983\end{array} \right.\\ \to \left\{ \begin{array}{l}x = – 1 – y\\ – 1 – y – 657y = – 1983\end{array} \right.\\ \to \left\{ \begin{array}{l} – 658y = – 1982\\x = – 1 – y\end{array} \right.\\ \to \left\{ \begin{array}{l}y = \dfrac{{991}}{{329}}\\x = – \dfrac{{1320}}{{329}}\end{array} \right.\\b)\left\{ \begin{array}{l}x + y = 1\\x – 657y = 1983\end{array} \right.\\ \to \left\{ \begin{array}{l}x = 1 – y\\1 – y – 567y = 1983\end{array} \right.\\ \to \left\{ \begin{array}{l}x = 1 – y\\ – 568y = 1982\end{array} \right.\\ \to \left\{ \begin{array}{l}y = – \dfrac{{991}}{{284}}\\x = \dfrac{{1275}}{{284}}\end{array} \right.\\c)\left\{ \begin{array}{l}x + y = – 1983\\x – 657y = – 1\end{array} \right.\\ \to \left\{ \begin{array}{l}x = – 1983 – y\\ – 1983 – y – 657y = – 1\end{array} \right.\\ \to \left\{ \begin{array}{l} – 658y = 1982\\x = – 1983 – y\end{array} \right.\\ \to \left\{ \begin{array}{l}y = – \dfrac{{991}}{{329}}\\x = – 1979,987842\end{array} \right.\\d)\left\{ \begin{array}{l}x + y = 1983\\x – 657y = 1\end{array} \right.\\ \to \left\{ \begin{array}{l}x = 657y + 1\\657y + 1 + y = 1983\end{array} \right.\\ \to \left\{ \begin{array}{l}y = \dfrac{{991}}{{329}}\\x = 1979,987842\end{array} \right.\end{array}\) Bình luận
Đáp án:
a) \(\left\{ \begin{array}{l}
y = \dfrac{{991}}{{329}}\\
x = – \dfrac{{1320}}{{329}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left\{ \begin{array}{l}
x + y = – 1\\
x – 657y = – 1983
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = – 1 – y\\
– 1 – y – 657y = – 1983
\end{array} \right.\\
\to \left\{ \begin{array}{l}
– 658y = – 1982\\
x = – 1 – y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{991}}{{329}}\\
x = – \dfrac{{1320}}{{329}}
\end{array} \right.\\
b)\left\{ \begin{array}{l}
x + y = 1\\
x – 657y = 1983
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 – y\\
1 – y – 567y = 1983
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 – y\\
– 568y = 1982
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = – \dfrac{{991}}{{284}}\\
x = \dfrac{{1275}}{{284}}
\end{array} \right.\\
c)\left\{ \begin{array}{l}
x + y = – 1983\\
x – 657y = – 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = – 1983 – y\\
– 1983 – y – 657y = – 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
– 658y = 1982\\
x = – 1983 – y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = – \dfrac{{991}}{{329}}\\
x = – 1979,987842
\end{array} \right.\\
d)\left\{ \begin{array}{l}
x + y = 1983\\
x – 657y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 657y + 1\\
657y + 1 + y = 1983
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{991}}{{329}}\\
x = 1979,987842
\end{array} \right.
\end{array}\)