Dwa równania z symbolem Newtona.

[jetix16]

prosze o rozwiazanie tych dwoch rownan..
z gory dzieki
\(\displaystyle{ {n\choose n-2}}\)=6
\(\displaystyle{ {n\choose 3}}\)-\(\displaystyle{ {n\choose 2}}\)=0

[soku11]

1.
\(\displaystyle{ n>n-2\ \Rightarrow\ 0>-2\ \Rightarrow\ n\in\mathbb{N}\\
\frac{n!}{(n-2)!2!}=6\\
\frac{(n-2)!(n-1)n}{(n-2)!2}=6\\
(n-1)n=12\\
...}\)

2.
\(\displaystyle{ n>3\ n>2\ \Rightarrow\ n>3\\
\frac{n!}{3!(n-3)!}-\frac{n!}{2!(n-2)!}=0 \\
\frac{(n-3)!(n-2)(n-1)n}{6(n-3)!}-\frac{(n-2)!(n-1)n}{2(n-2)!}=0 \\
\frac{(n-2)(n-1)n}{6}-\frac{(n-1)n}{2}=0 \\
(n-2)(n-1)n-3(n-1)n=0 \\
n(n-1)[(n-2)-3]=0\\
n(n-1)(n-5)=0\\
...}\)

POZDRO

[jetix16]

dzieki wielkie !