Question

calcule o valor de x. Sendo x!.(x+1)!/(x-1)!.x!=20

Answer 1

Resposta:

$x^{2} + x$÷$x^{2} -x=20$

$x^{3}$÷$x=20$

$x^{2} =20$

$x=\sqrt{20}$

$x=4,47213595499957...$

Answer 2

Resposta:

X=4

Explicação passo-a-passo:

$\displaystyle \frac{x!(x+1)!}{(x-1)!x!} =20\\\\\frac{\diagup\!\!\!\!x!(x+1)!}{(x-1)!\diagup\!\!\!\!x!} =20\\\\\frac{(x+1).x!}{(x-1)!} =20\\\\\frac{(x+1).x.(x-1)!}{(x-1)!} =20\\\\\frac{(x+1).x.\diagup\!\!\!\!\!\!\!(x-\diagup\!\!\!\!\!\!\!1)!}{\diagup\!\!\!\!\!\!\!(x-\diagup\!\!\!\!\!\!\!1)!} =20\\\\x(x+1)=20\\x^2+x-20=0$

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+x-20=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=1~e~c=-20\\\\\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(1)(-20)=1-(-80)=81\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{81}}{2(1)}=\frac{-1-9}{2}=\frac{-10}{2}=-5\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{81}}{2(1)}=\frac{-1+9}{2}=\frac{8}{2}=4\\\\S=\{-5,~4\}$

Para x= -5

Substituir na expressão:

$\displaystyle \frac{x!(x+1)!}{(x-1)!x!} =20\\\\\displaystyle \frac{(-5)!(-5+1)!}{(-5-1)!(-5)!} =20$

Não existe fatorial de um número negativo (-5)!, logo essa resposta não serve:

Para x=4

$\displaystyle \frac{x!(x+1)!}{(x-1)!x!} =20\\\\\frac{4!(4+1)!}{(4-1)!4!} =20\\\\\frac{5!}{3!} =20\\\\\frac{5.4.\diagup\!\!\!\!3!}{\diagup\!\!\!\!3!} =20\\20=20~~(VERDADEIRO)$