((6n^2-4n+2)/(12n^2+4n-6))-(1/2) kann das jemand eben mit zwischen schritten vollständig und ausführlich ausrechnen danke.
+14903 ((6n^2-4n+2)/(12n^2+4n-6))-(1/2) ausrechnen.
Hallo Gast!
\(\frac{6n^2-4n+2}{12n^2+4n-6}-\frac{1}{2}\)
\( =\frac{3n^2-2n+1}{6n^2+2n-3}-\frac{1}{2}\\ =\frac{2\cdot (3n^2-2n+1)-(6n^2+2n-3)}{(6n^2+2n-3)\cdot 2}\\ =\frac{3n^2-2n+1-3n^2-n+\frac{3}{2}}{6n^2+2n-3}\)
\(=\frac{-3n+\frac{5}{2}}{6n^2+2n-3}\\ =\frac{5-6n}{(6n^2+2n-3)\cdot2}\\ \color{blue}=\frac{5-6n}{12n^2+4n-6}\)
!
