Löse nach y auf

Löse nach y auf

Hallo mathefan!

 $\large \color{BrickRed}{y-m \over m-n}-{y \over m}+{m+y \over m+n}={m^2-ny \over mn}\\ $

$\large \frac{(y-m)(m+n)mn-yn(m-n)(m+n)+(m+y)mn(m-n)}{mn(m-n)(m+n)}=\frac{(m^2-ny)(m-n)(m+n)}{mn(m-m)(m+n)}$

$(y-m)(m+n)mn-yn(m-n)(m+n)+(m+y)mn(m-n)\\ =(m^2-ny)(m-n)(m+n)$

$(ym+yn-m^2-mn)mn-(m^2-n^2)yn+(m^2-mn+ym-yn)mn\\ =(m^2-n^2)(m^2-ny)$

$ym^2n+ymn^2-(m^2-n^2)yn+m^3n-m^2n^2+ym^2n-yn^2m\\ =m^4-m^2ny-n^2m^2+n^3y\\ .\\ ym^2n+ymn^2-(m^2-n^2)yn+ym^2n-yn^2m+m^2ny-n^3y\\ =m^4-n^2m^2-m^3n+m^2n^2$

$y\cdot (m^2n+mn^2-m^2n+n^3+m^2n-n^2m+m^2n-n^3)\\ =m^4-n^2m^2-m^3n+m^2n^2$

$\large y =\frac{m^4-n^2m^2-m^3n+m^2n^2}{m^2n+mn^2-m^2n+n^3+m^2n-n^2m+m^2n-n^3}\\ \large y=\frac{m^4-m^3n}{2m^2n}$

$\large \color{blue}y=\frac{m^2-mn}{2n}=\frac{m(m-n)}{2n}$

  !

Löse nach y auf