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[tex]\sf\sqrt{x+1}=x-1\\\sf(\sqrt{x+1})^2=(x-1)^2\\\sf x+1=x^2-2x+\diagup\!\!\!1\\\sf x^2-2x+1=x+\diagup\!\!\!1\\\sf x^2-2x-x=0\\\sf x^2-3x=0\\\sf x\cdot(x-3)=0\\\sf x=0\\\sf x-3=0\\\sf x=3\\\underline{\rm verificac_{\!\!,}\tilde ao:}\\\sf para~x=0\\\sf \sqrt{0+1}=0-1\\\sf\sqrt{1}=-1\\\sf 1=-1 (absurdo!)\implies x=0~n\tilde ao~\acute e~ra\acute iz.\\\sf para~x=3\\\sf \sqrt{3+1}=3-1\\\sf\sqrt{4}=2\\\sf 2=2\implies x=3~\acute e~ra\acute iz.\\\huge\boxed{\boxed{\boxed{\boxed{\sf S=\{3\}}}}}\blue{\checkmark}[/tex]
[tex]\sf\sqrt{4-5x}=x-2\\\sf(\sqrt{4-5x})^2=(x-2)^2\\\sf\diagdown\!\!\!4-5x=x^2-4x+\diagdown\!\!\!4\\\sf x^2-4x+5x=0\\\sf x^2+x=0\\\sf x\cdot(x+1)=0\\\sf x=0\\\sf x+1=0\\\sf x=-1\\\underline{\rm verificac_{\!\!,}\tilde ao:}\\\sf para~x=0\\\sf \sqrt{4-5\cdot0}=0-2\\\sf\sqrt{4}=-2\\\sf 2=-2(absurdo!)\implies x=0~n\tilde ao~\acute e~ra\acute iz.\\\sf para~x=-1\\\sf \sqrt{4-5\cdot(-1)}=-1-2\\\sf\sqrt{4+5}=-3\\\sf\sqrt{9}=-3\\\sf 3=-3\implies x=-1~n\tilde ao~\acute e~soluc_{\!\!,}\tilde ao\\\sf portanto[/tex]
[tex]\huge\boxed{\boxed{\boxed{\boxed{\sf S=\varnothing}}}}[/tex]
