[tex]\it a)\ 2x+1=\sqrt2 \Rightarrow 2x=-1+\sqrt2 \Rightarrow x=\dfrac{-1+\sqrt2}{2} \Rightarrow A=\Big\{\dfrac{-1+\sqrt2}{2}\ \Big\}\\ \\ \\ b)\ x^2=5 \Rightarrow \sqrt{x^2}=\sqrt5 \Rightarrow |x|=\sqrt5 \Rightarrow x=\pm\sqrt5 \Rightarrow B=\{-\sqrt5,\ \sqrt5\}\\ \\ \\ c)\ x^2\leq0\ \ \ \ \ (1)\\ \\ Dar,\ x^2\geq0,\ \forall\ x\in\mathbb{R}\ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow x^2=0 \Rightarrow x=0 \Rightarrow C=\{0\}\\ \\ \\ d)\ \ |x|=3 \Rightarrow x=\pm3 \Rightarrow D=\{-3,\ \ 3\}[/tex]
