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Question-172631




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

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