Matemática, perguntado por dudaelala, 1 ano atrás

Simpiflique as expressões

48!+49!
_______
50!


n!
___
(n+1)!


(n+3)! .(n-1)!
____________
(n-2)! (n+2)!

Soluções para a tarefa

Respondido por MATHSPHIS
1
\boxed{\frac{48!+49!}{50!}=\frac{48!+49.48!}{50*49*48!}=\frac{48!(1+49)}{50*49*48!}=\frac{1}{49}}

\boxed{\frac{n!}{(n+1)!}=\frac{n!}{(n+1)n!}=\frac{1}{n+1}}

\frac{(n+3)!(n-1)!}{(n-2)!(n+2)!}=\frac{(n+3)(n+2)(n+1)n(n-1)(n-2)!(n-1)!}{(n-2)!(n+2)(n+1)n(n-1)!}=\\
\\
=(n+3)(n-1)
Respondido por Usuário anônimo
0
a) \dfrac{48!+49!}{50!}=\dfrac{48!\cdot(1+49)}{50\cdot49\cdot48!}=\dfrac{50}{50\cdot49}=\dfrac{1}{49}


b) \dfrac{n!}{(n+1)!}=\dfrac{n!}{(n+1)n!}=\dfrac{1}{n+1}

c) \dfrac{(n+3)!\cdot(n-1)!}{(n-2)!\cdot(n+2)!}=\dfrac{(n+3)(n+2)!\cdot(n-1)(n-2)!}{(n-2)!\cdot(n+2)!}=(n+3)(n-1)
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