Question

Simplifiquem as expressões abaixo:
(obs: depois de cada número tem "!" esse sinal de fatorial) com calculos pls

a)
$\frac{100 + 99}{99}$
b)
$\frac{4 + 5}{3}$
​

Answer 1

$\large\boxed{\begin{array}{l}\\\underline{\rm Resoluc_{\!\!,}\tilde ao\,para\,vers\tilde ao\,da\,web }\\\\\rm a)~\sf\dfrac{100!+99!}{99!}=\dfrac{100\cdot99!+99!}{99!}\\\\\sf=\dfrac{\diagdown\!\!\!\!\!\!99\!!(100+1)}{\diagdown\!\!\!\!\!\!99\!!}=100+1=101\checkmark\end{array}}$$\large\boxed{\begin{array}{l}\underline{\rm Resoluc_{\!\!,}\tilde ao\,para\,vers\tilde ao\,do\,app}\\\\\rm a)~\sf\dfrac{100!+99!}{99!}=\dfrac{100\cdot99!+99!}{99!}\\\\\sf=\dfrac{\cancel{99!}(100+1)}{\cancel{99!}}=100+1=101\blue{\checkmark}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Resoluc_{\!\!,}\tilde ao\,para\,vers\tilde ao\,da\,web}\\\\\rm b)~\sf\dfrac{4!+5!}{3!}=\dfrac{4\cdot3!+5\cdot4\cdot3!}{3!}\\\\\sf\dfrac{\diagup\!\!\!\!\!3!(4+20)}{\diagup\!\!\!\!3!}=4+20=24\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Resoluc_{\!\!,}\tilde ao\,para\,vers\tilde ao\,do\,app}\\\\\rm b)~\sf\dfrac{4!+5!}{3!}=\dfrac{4\cdot3!+5\cdot4\cdot3!}{3!}\\\\\sf=\dfrac{\cancel{3!}(4+20)}{\cancel{3!}}=4+20=24\blue{\checkmark}\end{array}}$