Toán Lớp 10: giải hệ
x ²-3.y ²+2xy-2x-10y+4=0
x ²+y ²=10
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 10: giải hệ
x ²-3.y ²+2xy-2x-10y+4=0
x ²+y ²=10
Comments ( 1 )
\left\{ \begin{array}{l}
{x^2} – 3{y^2} + 2xy – 2x – 10y + 4 = 0\\
{x^2} + {y^2} = 10
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} – 3{y^2} + 2xy – 2x – 10y + 4 = 0\left( 1 \right)\\
{x^2} = 10 – {y^2}\left( 2 \right)
\end{array} \right.\\
\left( 2 \right) \to \left( 1 \right):\left( {10 – {y^2}} \right) – 3{y^2} + 2xy – 2x – 10y + 4 = 0\\
\Leftrightarrow 10 – 4{y^2} + 2xy – 2x – 10y + 4 = 0\\
\Leftrightarrow 10 – 10y + 2x\left( {y – 1} \right) – 4{y^2} + 4 = 0\\
\Leftrightarrow 10\left( {1 – y} \right) + 2x\left( {y – 1} \right) + 4\left( {1 – {y^2}} \right) = 0\\
\Leftrightarrow 10\left( {1 – y} \right) + 2x\left( {y – 1} \right) – 4\left( {{y^2} – 1} \right) = 0\\
\Leftrightarrow – 4\left( {y – 1} \right)\left( {y + 1} \right) + 2x\left( {y – 1} \right) – 10\left( {y – 1} \right) = 0\\
\Leftrightarrow \left( {y – 1} \right)\left[ { – 4\left( {y + 1} \right) + 2x – 10} \right] = 0\\
\Leftrightarrow \left( {y – 1} \right)\left( {2x – 4y – 14} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y = 1\\
x – 2y – 7 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 1\left( 3 \right)\\
x = 2y + 7\left( 4 \right)
\end{array} \right.\\
\left( 3 \right) \to \left( 2 \right):{x^2} = 10 – 1 = 9 \Rightarrow {x^2} = 9 \Rightarrow x = \pm 3\\
\left( 4 \right) \to \left( 2 \right):{\left( {2y + 7} \right)^2} = 10 – {y^2}\\
\Leftrightarrow 4{y^2} + 28y + 49 = 10 – {y^2}\\
\Leftrightarrow 5{y^2} + 28y + 39 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y = – 3 \Rightarrow x = 1\\
y = – \dfrac{{13}}{5} \Rightarrow x = \dfrac{9}{5}
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left( {3;1} \right),\left( { – 3;1} \right),\left( {1; – 3} \right),\left( {\dfrac{9}{5}; – \dfrac{{13}}{5}} \right)
\end{array}$