Giải hệ phương trình: 2x^2-y^2=-14 5x-y=9

Đáp án:

 \(\left( {x;y} \right) \in \left\{ {\left( {\dfrac{{67}}{{23}};\dfrac{{128}}{{23}}} \right);\left( {1; – 4} \right)} \right\}\)

Giải thích các bước giải:

$\begin{array}{l}
\left\{ \begin{array}{l}
2{x^2} – {y^2} =  – 14\\
5x – y = 9
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2{x^2} – {y^2} =  – 14\\
y = 5x – 9
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2{x^2} – {\left( {5x – 9} \right)^2} =  – 14\\
y = 5x – 9
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2{x^2} – \left( {25{x^2} – 90x + 81} \right) =  – 14\\
y = 5x – 9
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2{x^2} – 25{x^2} + 90x – 81 =  – 14\\
y = 5x – 9
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
 – 23{x^2} + 90x – 67 = 0\\
y = 5x – 9
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = \dfrac{{67}}{{23}}\\
x = 1
\end{array} \right.\\
y = 5x – 9
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = \dfrac{{67}}{{23}}\\
y = \dfrac{{128}}{{23}}
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 1\\
y =  – 4
\end{array} \right.
\end{array} \right.\\
Vay\,\,\left( {x;y} \right) \in \left\{ {\left( {\dfrac{{67}}{{23}};\dfrac{{128}}{{23}}} \right);\left( {1; – 4} \right)} \right\}
\end{array}$