Question

Нужно выразить через и
уравнение:

Answer 1

$(\frac{f_{0}\rho_{1}}{\sqrt{(\omega_{0}^2-\rho_{1}^2)^2+4b^2\rho_{1}^2} } )^2=(\frac{f_{0}\rho_{2}}{\sqrt{(\omega_{0}^2-\rho_{2}^2)^2+4b^2\rho_{2}^2}})^2\\ \dfrac{f_{0}^2\rho_{1}^2}{{(\omega_{0}^2-\rho_{1}^2)^2+4b^2\rho_{1}^2} } =\dfrac{f_{0}^2\rho_{2}^2}{{(\omega_{0}^2-\rho_{2}^2)^2+4b^2\rho_{2}^2}}\\ f_{0}^2\rho_{1}^2({(\omega_{0}^2-\rho_{2}^2)^2+4b^2\rho_{2}^2})=f_{0}^2\rho_{2}^2((\omega_{0}^2-\rho_{1}^2)^2+4b^2\rho_{1}^2} })\\$

$\rho_{1}^2({\omega_{0}^2-\rho_{2}^2)^2}=\rho_{2}^2(\omega_{0}^2-\rho_{1}^2)^2}\\ \rho_{1}^2({\omega_{0}^4-2\omega_{0}^2\rho_{2}^2+\rho_{2}^4)}=\rho_{2}^2(\omega_{0}^4-2\omega_{0}^2\rho_{1}^2+\rho_{1}^4)}\\ \rho_{1}^2({\omega_{0}^4+\rho_{2}^4)}=\rho_{2}^2({\omega_{0}^4+\rho_{1}^4)}\\ \omega_{0}^4(\rho_{1}^2-\rho_{2}^2)=\rho_{1}^2\rho_{2}^2(\rho_{1}^2-\rho_{2}^2)\\ \omega_{0}=\pm\sqrt{\rho_{1}\rho_{2}}$

Подразумевается, что $f_0\neq 0,\:\rho_1\neq\rho_2\neq0$