Question

40 баллов
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Answer 1

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Answer 2

$\frac{x}{y} \left \{ {{2x^2-3xy+2y^2=4} \atop {2x^2+3y^2=14}} \right.\; \ominus \; \left \{ {{2x^2-3xy+2y^2=4\; |\cdot 10} \atop {y^2+3xy=10\; \; |\cdot (-4)}} \right.\; \oplus \; \left \{ {{2x^2-3xy+2y^2=4} \atop {20x^2-42xy+16y^2=0|:2}} \right.\\\\10x^2-21xy+8y^2=0\; |:y^2\ne 0\\\\10(\frac{x}{y})^2-21\frac{x}{y}+8=0\\\\t=\frac{x}{y}\; ,\; \; \; 10t^2-21t+8=0\; \; ,\; \; D=121\; ,\; t_1=\frac{1}{2}\; ,\; \; t_2=\frac{8}{5}\\\\a)\; \; \frac{x}{y}=\frac{1}{2}\; \; \to \; \;  y=2x\; ,\; \; 2x^2-3x\cdot 2x+2\cdot 4x^2=4\; ,\; \; 4x^2=4$

$x^2=1\; \; \to \; \; \; x_1=1\; ,\; \; x_2=-1\\\\y_1=2\; ,\; y_2=-2\\\\b)\; \; \frac{x}{y}=\frac{8}{5}\; \; \to \; \; y=\frac{5x}{8}\; \; ,\; \; 2x^2-\frac{15x^2}{8}+\frac{50x^2}{64}=4\; ,\; \; \frac{58x^2}{64}=4\; ,\\\\x^2=\frac{256}{58}=\frac{128}{29}\; ,\; \; x_1=\frac{8\sqrt2}{\sqrt{29}}\; \; ,\; \; x_2=-\frac{8\sqrt2}{\sqrt{29}}\\\\y_1=\frac{5\sqrt2}{\sqrt{29}}\; \; ,\; \; y_2=-\frac{5\sqrt2}{\sqrt{29}}\\\\Otvet:\; \; (1,2)\; ,\; (-1,-2)\; ,\; (\, \frac{8\sqrt2}{\sqrt{29}}\, ,\, \frac{5\sqrt2}{\sqrt{29}}\, )\; ,\; \;  (-\frac{8\sqrt2}{\sqrt{29}},-\frac{5\sqrt2}{\sqrt{29}}\, )\; .$