Tìm x, y theo beta :3x^2 +y^2+4xy+4x+2y+5=0 28/08/2021 Bởi Sarah Tìm x, y theo beta :3x^2 +y^2+4xy+4x+2y+5=0
Đáp án: $\begin{array}{l}3{x^2} + {y^2} + 4xy + 4x + 2y + 5 = 0\\ \Rightarrow 4{x^2} + {y^2} + 1 + 4xy + 4x + 2y – {x^2} + 4 = 0\\ \Rightarrow {\left( {2x} \right)^2} + {y^2} + 1 + 2.2.y + 2.2x + 2.y.1 – {x^2} + 4 = 0\\ \Rightarrow {\left( {2x + y + 1} \right)^2} – {x^2} = – 4\\ \Rightarrow \left( {2x + y + 1 – x} \right)\left( {2x + y + 1 + x} \right) = – 4\\ \Rightarrow \left( {x + y + 1} \right)\left( {3x + y + 1} \right) = – 4 = \left( { – 1} \right).4 = 1.\left( { – 4} \right) = \left( { – 4} \right).1 = 4.\left( { – 1} \right) = 2.\left( { – 2} \right) = \left( { – 2} \right).2\\ \Rightarrow \left[ \begin{array}{l}x = \frac{5}{2};y = \frac{9}{2}\\x = \frac{{ – 5}}{2};y = \frac{5}{2}\\x = \frac{5}{2};y = \frac{{ – 15}}{2}\\x = \frac{{ – 5}}{2};y = \frac{{11}}{2}\\x = – 2;y = 3\\x = 2;y = – 5\end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
3{x^2} + {y^2} + 4xy + 4x + 2y + 5 = 0\\
\Rightarrow 4{x^2} + {y^2} + 1 + 4xy + 4x + 2y – {x^2} + 4 = 0\\
\Rightarrow {\left( {2x} \right)^2} + {y^2} + 1 + 2.2.y + 2.2x + 2.y.1 – {x^2} + 4 = 0\\
\Rightarrow {\left( {2x + y + 1} \right)^2} – {x^2} = – 4\\
\Rightarrow \left( {2x + y + 1 – x} \right)\left( {2x + y + 1 + x} \right) = – 4\\
\Rightarrow \left( {x + y + 1} \right)\left( {3x + y + 1} \right) = – 4 = \left( { – 1} \right).4 = 1.\left( { – 4} \right) = \left( { – 4} \right).1 = 4.\left( { – 1} \right) = 2.\left( { – 2} \right) = \left( { – 2} \right).2\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{5}{2};y = \frac{9}{2}\\
x = \frac{{ – 5}}{2};y = \frac{5}{2}\\
x = \frac{5}{2};y = \frac{{ – 15}}{2}\\
x = \frac{{ – 5}}{2};y = \frac{{11}}{2}\\
x = – 2;y = 3\\
x = 2;y = – 5
\end{array} \right.
\end{array}$