30 điểm nà giúp mình từ g đến i nha love♥♥ November 13, 2020 by Bơ 30 điểm nà giúp mình từ g đến i nha love♥♥
Đáp án: h. \(\left\{ \begin{array}{l}y = 5\\x = 6\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}g.\left\{ \begin{array}{l}x = 5 – 2y\\{\left( {5 – 2y + 3} \right)^2} + {\left( {y – 1} \right)^2} = {\left( {5 – 2y} \right)^2} + {y^2}\left( 1 \right)\end{array} \right.\\\left( 1 \right) \to {\left( {8 – 2y} \right)^2} + {y^2} – 2y + 1 = 25 – 20y + 4{y^2} + {y^2}\\ \to 64 – 32y + 4{y^2} + {y^2} – 2y + 1 = 25 – 20y + 4{y^2} + {y^2}\\ \to 14y = 40\\ \to y = \dfrac{{20}}{7}\\ \to x = – \dfrac{5}{7}\\i.\left\{ \begin{array}{l}y = 5 – 3x\\{\left( {x – 1} \right)^2} – 5 + 3x + y = {x^2} + 10\left( 2 \right)\end{array} \right.\\\left( 2 \right) \to {x^2} – 2x + 1 – 5 + 3x + y = {x^2} + 10\\ \to x = 14\\ \to y = – 37\\h.\left\{ \begin{array}{l}xy – y + x – 1 = xy\\2x – 3y = – 3\end{array} \right.\\ \to \left\{ \begin{array}{l}x – y = 1\\2x – 3y = – 3\end{array} \right.\\ \to \left\{ \begin{array}{l}x = 1 + y\\2\left( {1 + y} \right) – 3y = – 3\end{array} \right.\\ \to \left\{ \begin{array}{l}x = 1 + y\\2 + 2y – 3y = – 3\end{array} \right.\\ \to \left\{ \begin{array}{l}y = 5\\x = 6\end{array} \right.\end{array}\) Reply
Đáp án:
h. \(\left\{ \begin{array}{l}
y = 5\\
x = 6
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
g.\left\{ \begin{array}{l}
x = 5 – 2y\\
{\left( {5 – 2y + 3} \right)^2} + {\left( {y – 1} \right)^2} = {\left( {5 – 2y} \right)^2} + {y^2}\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to {\left( {8 – 2y} \right)^2} + {y^2} – 2y + 1 = 25 – 20y + 4{y^2} + {y^2}\\
\to 64 – 32y + 4{y^2} + {y^2} – 2y + 1 = 25 – 20y + 4{y^2} + {y^2}\\
\to 14y = 40\\
\to y = \dfrac{{20}}{7}\\
\to x = – \dfrac{5}{7}\\
i.\left\{ \begin{array}{l}
y = 5 – 3x\\
{\left( {x – 1} \right)^2} – 5 + 3x + y = {x^2} + 10\left( 2 \right)
\end{array} \right.\\
\left( 2 \right) \to {x^2} – 2x + 1 – 5 + 3x + y = {x^2} + 10\\
\to x = 14\\
\to y = – 37\\
h.\left\{ \begin{array}{l}
xy – y + x – 1 = xy\\
2x – 3y = – 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x – y = 1\\
2x – 3y = – 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 + y\\
2\left( {1 + y} \right) – 3y = – 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 + y\\
2 + 2y – 3y = – 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 5\\
x = 6
\end{array} \right.
\end{array}\)