Question

can you solve this question​

Answer 1

Відповідь:

Пояснення:

Answer 2

Відповідь:

(2;1)    &    (-2;-1)

Пояснення:

$\left \{ {{\frac{5}{x^2+xy} +\frac{4}{y^2+xy} =\frac{13}{6} } \atop {{\frac{8}{x^2+xy} -\frac{1}{y^2+xy} =1}}/*4 \right. \\ \left \{ {{\frac{5}{x^2+xy} +\frac{4}{y^2+xy} =\frac{13}{6} } \atop {{\frac{32}{x^2+xy} -\frac{4}{y^2+xy} =4}} \right.\\\frac{5}{x^2+xy}+\frac{4}{y^2+xy}+\frac{32}{x^2+xy} -\frac{4}{y^2+xy} =\frac{13}{6}+4\\\frac{5}{x^2+xy}+\frac{32}{x^2+xy}=\frac{37}{6} \\\frac{37}{x^2+xy}=\frac{37}{6}\\x^2+xy=6\\\frac{5}{6} +\frac{4}{y^2+xy} =\frac{13}{6}$

$\frac{4}{y^2+xy} =\frac{13}{6} -\frac{5}{6} \\\frac{4}{y^2+xy} =\frac{8}{6} =\frac{4}{3} \\y^2+xy=3\\\left \{ {{x^2+xy=6} \atop {y^2+xy=3}} \right. \\y=\frac{6-x^2}{x} \\(\frac{6-x^2}{x})^2+6-x^2=3\\\frac{36-12x^2+x^4}{x^2} -\frac{x^4}{x^2} =-3\\\frac{36-12x^2}{x^2} =-3\\36-12x^2=-3x^2\\9x^2=36\\x^2=4\\x_1=2\\x_2=-2\\y_1=\frac{6-x_1^2}{x_1}=\frac{6-2^2}{2}=1 \\y_2=\frac{6-x_2^2}{x_2}=\frac{6-(-2)^2}{-2}=-1$