Toán Lớp 9: $\left \{ {{x^2-xy+y^2=1} \atop {x^2+xy+2y^2=4}} \right.$
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Toán Lớp 9: $\left \{ {{x^2-xy+y^2=1} \atop {x^2+xy+2y^2=4}} \right.$
giúp với ạ
Comments ( 1 )
\left\{ \begin{array}{l}
{x^2} – xy + {y^2} = 1\left( 1 \right)\\
{x^2} + xy + 2{y^2} = 4\left( 2 \right)
\end{array} \right.\\
4.\left( 1 \right) – \left( 2 \right) \Rightarrow 3{x^2} – 5xy + 2{y^2} = 0\\
\Leftrightarrow \left( {x – y} \right)\left( {3x – 2y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = y\left( 3 \right)\\
3x = 2y\left( 4 \right)
\end{array} \right.\\
\left( 3 \right) \to \left( 1 \right):{x^2} – {x^2} + {x^2} = 1\\
\Rightarrow x = \pm 1 \Rightarrow y = \pm 1\\
\left( 4 \right) \to \left( 1 \right):{x^2} – x.\dfrac{{3x}}{2} + {\left( {\dfrac{{3x}}{2}} \right)^2} = 1\\
\Leftrightarrow {x^2} – \dfrac{{3{x^2}}}{2} + \dfrac{{9{x^2}}}{4} = 1\\
\Leftrightarrow 4{x^2} – 6{x^2} + 9{x^2} = 4\\
\Leftrightarrow 7{x^2} = 4\\
\Leftrightarrow x = \pm \dfrac{{2\sqrt 7 }}{7} \Rightarrow y = \pm \dfrac{{3\sqrt 7 }}{7}\\
\Rightarrow \left( {x;y} \right) = \left( {1;1} \right)\left( { – 1;1} \right),\left( {\dfrac{{2\sqrt 7 }}{7};\dfrac{{3\sqrt 7 }}{7}} \right)\left( { – \dfrac{{2\sqrt 7 }}{7}; – \dfrac{{3\sqrt 7 }}{7}} \right)
\end{array}$