Question

Ajudem urgenteee!!
obtenha o valor de x na equação fatorial ....
(x+1)!+x!
______ = 6x
(x-1)!

Answer 1

               $\frac{(x+1)!+x!}{(x-1)!} =6x \\ \\ \frac{(x+1)(x)(x-1)!+x(x-1)!}{(x-1)!} =6x \\ \\ \frac{(x-1)![(x+1)(x)+x]}{(x-1)!} =6x \\ \\ x^2+x+x=6x \\ \\ x^2-4x=0 \\ \\ x(x-4)=0 \\ \\ x1=0 \\ x2=4 \\ \\ s$

                                           S = { 0, 4 }

Answer 2

$\dfrac{(x+1)!+x!}{(x-1)!}=6x\qquad\qquad\qquad\begin{matrix}\begin{matrix}\mathrm{Condi}\text{\c{c}}\mathrm{\tilde{a}o\ de\ exist\hat{e}ncia\ do\ fatorial:}\end{matrix}\\\\\begin{Bmatrix}\text{a)}\qquad{x}+1\ge0\longrightarrow{x}\ge-1\\\\\text{b)}\qquad{x}-1\ge0\longrightarrow{x}\ge1\end{matrix}\\\\\begin{matrix}\text{C.E.}=\{x\in\mathbb{R}\mid{x}\ge1\}\end{matrix}\end{matrix}\\\\\\\dfrac{(x+1)!}{(x-1)!}+\dfrac{x!}{(x-1)!}=6x\\\\\\\dfrac{(x+1)(x)(x-1)!}{(x-1)!}+\dfrac{(x)(x-1)!}{(x-1)!}=6x\\\\\\(x+1)(x)+(x)=6x\\\\(x^2+x)+(x)=6x\\\\x^2+x+x=6x\\\\x^2+2x=6x\\\\x^2+2x-6x=0\\\\x^2-4x=0\\\\x(x-4)=0\\\\(x+0)(x-4)=0\left\langle\begin{matrix}x+0=0\longrightarrow{x}'=0\notin\text{C.E.}\\\\x-4=0\longrightarrow\boxed{{x}''=4}\in\text{C.E.}\end{matrix}\right\\\\\\\boxed{\boxed{S=\{4\}}}$