Toán Lớp 10: 3/(x ^ 2) = 2x + y, \qquad 3/(y ^ 2) = 2y + x
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Toán Lớp 10: 3/(x ^ 2) = 2x + y, \qquad 3/(y ^ 2) = 2y + x
Comments ( 1 )
Dkxd:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{3}{{{x^2}}} = 2x + y\\
\dfrac{3}{{{y^2}}} = 2y + x
\end{array} \right.\\
\Leftrightarrow \dfrac{3}{{{x^2}}} – \dfrac{3}{{{y^2}}} = 2x + y – 2y – x\\
\Leftrightarrow \dfrac{{3.\left( {{y^2} – {x^2}} \right)}}{{{x^2}{y^2}}} = x – y\\
\Leftrightarrow \dfrac{{3.\left( {y + x} \right)\left( {y – x} \right)}}{{{x^2}{y^2}}} + y – x = 0\\
\Leftrightarrow \left( {y – x} \right).\left( {\dfrac{{3y + 3x}}{{{x^2}{y^2}}} + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = y\\
\dfrac{{3y + 3x}}{{{x^2}{y^2}}} = – 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = y\\
3y + 3x = – {x^2}{y^2}\left( {ktm} \right)
\end{array} \right.\\
\Leftrightarrow x = y\\
\Leftrightarrow \dfrac{3}{{{x^2}}} = 2x + x\\
\Leftrightarrow \dfrac{3}{{{x^2}}} = 3x\\
\Leftrightarrow 3{x^3} = 3\\
\Leftrightarrow 3\left( {{x^3} – 1} \right) = 0\\
\Leftrightarrow {x^3} = 1\\
\Leftrightarrow x = 1\left( {tm} \right)\\
\Leftrightarrow x = y = 1\\
Vậy\,x = y = 1
\end{array}$