Нужно выразить $omega_{0}$ через $rho_{1}$ и $rho_{2}$
Уравнение:
$frac{f_{0}rho_{1}}{sqrt{(omega_{0}^2-rho_{1}^2)^2+4b^2rho_{1}^2} } =frac{f_{0}rho_{2}}{sqrt{(omega_{0}^2-rho_{2}^2)^2+4b^2rho_{2}^2}}$

Ответ

Автор - igorShap

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

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

Подразумевается, что $f_0neq 0,:rho_1neqrho_2neq0$

Нужно выразить через и Уравнение:

Ответ

Нужно выразить $omega_{0}$ через $rho_{1}$ и $rho_{2}$
Уравнение:
$frac{f_{0}rho_{1}}{sqrt{(omega_{0}^2-rho_{1}^2)^2+4b^2rho_{1}^2} } =frac{f_{0}rho_{2}}{sqrt{(omega_{0}^2-rho_{2}^2)^2+4b^2rho_{2}^2}}$