Question

Найдите х: 1/(m+n)+(m+n)/x=1/(m-n)+(m-n)/x *
1) (m+n)²
2) другой ответ
3) m² - n²
4) (m-n)²

Answer 1

$\dfrac1{m+n}+\dfrac{m+n}x=\dfrac1{m-n}+\dfrac{m-n}x\\\\\\\dfrac{x+(m+n)^2}{x(m+n)}=\dfrac{x+(m-n)^2}{x(m-n)}\\\\\\\dfrac{x+m^2+2mn+n^2}{x(m+n)}-\dfrac{x+m^2-2mn+n^2}{x(m-n)}=0\\\\\\\dfrac{(m-n)(x+m^2+2mn+n^2)-(m+n)(x+m^2-2mn+n^2)}{x(m+n)(m-n)}=0\\\\\\\dfrac{mx+m^3+2m^2n+mn^2-nx-m^2n-2mn^2-n^3}{x(m+n)(m-n)}-\\\\-\dfrac{mx+m^3-2m^2n+mn^2+nx+m^2n-2mn^2+n^3}{x(m+n)(m-n)}=0\\\\\\\dfrac{4m^2n-2nx-2m^2n-2n^3}{x(m+n)(m-n)}=0\\\\\\\dfrac{2m^2n-2nx-2n^3}{x(m+n)(m-n)}=0\\\\\\\dfrac{2n(m^2-x-n^2)}{x(m+n)(m-n)}=0$

$2n(m^2-x-n^2)=0\quad|:2\\\\n(m^2-x-n^2)=0\\{}\qquad\quad\Downarrow\\1)\ \ n=0;\ \ togda \ \ m^2-x\ \ ne\ \ opredeleno\ \ (libo =0;\ libo \ne0)\\\\2) \ \ n\ne 0;\ \ togda \ \ \ \begin{array}{lcl}m^2-x-n^2=0\\x=m^2-n^2\end{array}$

Ответ:  3;  при  $n\ne0;\ \ x=m^2-n^2$