Question

Me ajudem por favor!!! Para hoje!!!
a)(x+3)!/(x+1) = 3!
b)(x+2)/x! = -8

Answer 1

(x+3)! / x+1 = 3!
(x+3) . (x+2) . (x+1)! / (x+1)! = 3!
(x+3)(x+2) = 3!
x² + 5x +6 = 6
x² + 5x = 0
x(x+5) = 0
x=0
x+5= 0
x=-5

(x+2)! / x! = -8
(x+2) . (x+1) . x! / x! = -8
(x+2)(x+1) = -8
x²+3x+2 = -8
x²+3x+2+8=0
x²+3x+10=0
(x+5)(x-2)=0
x+5=0
x=-5
x-2=0
x= 2

Answer 2

Olá, boa tarde! ☺

Prezado amigo (a), com base no enunciado acima, podemos compreender que:

a)

$\frac {(x+3)!}{x+1} = 3!$

$\frac {(x+3)\cdot (x+2)\cdot (x+1)!}{(x+1)!} = 3!$

$(x+3)(x+2) = 3!$

$x^2 + 5x +6 = 6$

$x^2 + 5x = 0$

$x(x+5) = 0$

$x=0$

$x+5= 0$

$\boxed {x=-5}$

b)

$\frac {(x+2)!}{x!} = -8(x+2)\cdot (x+1)\cdot \frac {x!}{x}! = -8$

$(x+2)(x+1) = -8$

$x^2+3x+2 = -8$

$x^2+3x+2+8=0$

$x^2+3x+10=0$

$(x+5)(x-2)=0$

$x+5=0$

$x=-5$

$x-2=0$

$\boxed {x= 2}$