[tex]a)\hat{2}x+\hat{1}=\hat{0}\\\hat{2}x=\hat{2}\\x=\hat{1} \text{(singura solutie care convine)}\\b)x^2+\hat{1}=\hat{0}\\ x^2=\hat{4}\\\text{Observam ca }x\in \{2,3\}\text{ sunt singurele solutii.}\\c)\hat{3}x^2-x+\hat{2}=\hat{0}\\\hat{3}x^2-x=\hat{2}\\\text{Din nou,observam ca solutia este }x\in \{ 1,2\}.\\d)x^3+\hat{2}x+\hat{3}=\hat{0}\\\text{Aici se poate descompune:}\\x^3+x^2-x^2-x+\hat{3}x+\hat{3}=\hat{0}\\x^2(x+\hat{1})-x(x+\hat{1})+\hat{3}(x+\hat{1})=\hat{0}[/tex]
[tex](x+\hat{1})(x^2-x+\hat{3})=\hat{0}\\\text{Distingem doua cazuri posibile:}\\i)x+\hat{1}=0\Rightarrow x=\hat{4}\\ii)x^2-x+\hat{3}=0,\text{ cu solutia }x\in \{2,4\}\\\text{Prin urmare ,solutia este }x\in \{2,4\}[/tex]
