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DK: X.Y khac 0 \( \left\{ \begin{array}{l} 1+xy=\frac{6xy}{2x+y}\\ \frac{x^{2}+y^{2}}{(xy)^{2}}.(1+xy)=8\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 1+xy=\frac{6xy}{2x+y} (1)\\ \frac{x^{2}+y^{2}}{(xy)^{2}}.(\frac{6xy}{2x+y})^{2}=8 (2)\end{array} \right. \) giai 2 :(2)$\Leftrightarrow 0,5x^{2}+3,5y^{2}-4xy=0$ (3) ta thay y=o khong thoa man pt .do do ta chia ca 2 ve cho $y^{2}\Rightarrow $ pt 3 tro thanh pt bac 2 an $\frac{x}{y}$ giai ra ta dc: TH1 :$\frac{x}{y}=7$ TH2: $\frac{x}{y}=1$ (den day tu giai nha em.thay vao pt 1 la OK)
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