Question

Пожалуйста помогите .......

Answer 1

$\left\{\begin{array}{ccc}\frac{x}{y}+\frac{y}{x}=\frac{34}{15}\; \; \\x^2+y^2=34\end{array}\right\; \; \left\{\begin{array}{ccc}\frac{x^2+y^2}{xy}=\frac{34}{15}\quad \\x^2+y^2=34\end{array}\right\; \; \left\{\begin{array}{ccc}\frac{34}{xy}=\frac{34}{15}\qquad \\x^2+y^2=34\end{array}\right\; \; \left\{\begin{array}{ccc}xy=15\qquad \\x^2+y^2=34\end{array}\right$

$\left\{\begin{array}{ccc}y=\frac{15}{x}\qquad \quad \\x^2+\frac{225}{x^2}=34\end{array}\right\; \; \left\{\begin{array}{ccc}y=\frac{15}{x}\qquad \qquad \qquad \qquad \quad \\x^4-34x^2+225=0\; ,\; x\ne 0\end{array}\right\\\\\\x^4-34x^2+225=0\; ,\; \; D/4=64\; ,\; \; x^2=9\; \; ,\; \; x^2=25\\\\x_{1,2}=\pm 3\; \; ,\; \; x_{3,4}=\pm 5\\\\y_{1,2}=\pm 5\; \; ,\; \; y_{3,4}=\pm 3\\\\Otvet:\; \; (-3,-5)\; ,\; (3,5)\; ,\; (-5,-3)\; ,\; (5,3)\; .$