Question Number 203676 by 073 last updated on 25/Jan/24
Answered by Frix last updated on 25/Jan/24
$${x}\left({x}+\mathrm{2}\right){x}!=\mathrm{168}{x}^{\mathrm{2}} \\ $$$${x}=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right){x}!=\mathrm{168}{x} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right)!=\mathrm{168}=\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{7} \\ $$$${x}=\mathrm{5} \\ $$