Question Number 172631 by Mikenice last updated on 29/Jun/22
Commented by mr W last updated on 29/Jun/22
$${x}!=\left({x}+\mathrm{1}\right){x}\left({x}−\mathrm{1}\right) \\ $$$${x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)!=\left({x}+\mathrm{1}\right){x}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{2}\right)!={x}+\mathrm{1} \\ $$$${let}\:{u}={x}−\mathrm{2} \\ $$$$\Rightarrow{u}!={u}+\mathrm{3} \\ $$$$\Rightarrow{u}=\mathrm{3}\:\Rightarrow{x}=\mathrm{5}\:\checkmark \\ $$
Commented by Rasheed.Sindhi last updated on 29/Jun/22
$$\cap\boldsymbol{\mathrm{i}}\subset\in\:\boldsymbol{\mathrm{sir}}! \\ $$
Commented by Tawa11 last updated on 30/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$