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\({y-m \over m-n}-{y \over m}+{m+y \over m+n}={m^2-ny \over mn}\)

 15.03.2019
 #1
avatar+8576 
+2

Löse nach y auf

 

Hallo mathefan!

 

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

laugh  !

 15.03.2019
bearbeitet von asinus  16.03.2019
bearbeitet von asinus  16.03.2019
bearbeitet von asinus  16.03.2019
 #2
avatar+10536 
+1

Löse nach y auf

 16.03.2019

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