Question

Resolva as equações:

a) (x-11)!/ (x-3)! - (x-2)!/ (x-4)!= 14 - (x-3)! / (x-5)!

b) 8.(x+3)! / (x+4)! = 2

Answer 1

Olá, Estter.

$a)\,\,\frac{(x-1)!}{(x-3)!} - \frac{(x-2)!}{(x-4)!}= 14 - \frac{(x-3)!}{(x-5)!}\Rightarrow\\\\ \frac{(x-1)(x-2)(x-3)!}{(x-3)!} - \frac{(x-2)(x-3)(x-4)!}{(x-4)!}= 14 - \frac{(x-3)(x-4)(x-5)!}{(x-5)!}\Rightarrow\\\\ (x-1)(x-2)-(x-2)(x-3)=14-(x-3)(x-4)\Rightarrow\\\\ x^2-2x-x+2-(x^2-3x-2x+6)=14-(x^2-4x-3x+12)\Rightarrow\\\\ x^2-3x+2-x^2+5x-6=14-x^2+7x-12\Rightarrow\\\\ 2x-4=-x^2+7x+2\Rightarrow x^2-5x-6=0\Rightarrow\\\\ \Delta=(-5)^2-4\cdot1\cdot(-6)=25+24=49\Rightarrow\sqrt\Delta=7\Rightarrow$

$x=\frac{5\pm\sqrt\Delta}2\Rightarrow\,x=\frac{5\pm7}2\Rightarrow\boxed{x=6}\\\\\text{ (desprezamos }x = -1\text{, pois n\~ao existe fatorial de n\'umero negativo)}$


$b)\,\, \frac{8(x+3)!}{(x+4)!} = 2 \Rightarrow \\\\\frac{8(x+3)!}{(x+4)(x+3)!} = 2 \Rightarrow \\\\\frac{8}{(x+4)} = 2 \Rightarrow \\\\8=2(x+4)\Rightarrow \\\\ 8=2x+8\Rightarrow \\\\2x=0\Rightarrow \\\\\boxed{x=0}$