$\displaystyle \begin{array}{{>{\displaystyle}l}} b.\ \sqrt{x^{2} +2021} >x\sqrt{2020}( 1)\\ TH1:\ x< 0\Rightarrow ( 1) \ luôn\ đúng.\\ TH2:\ x\geqslant 0\\ ( 1) \Leftrightarrow x^{2} +2021 >2020x^{2}\\ \Leftrightarrow 2019x^{2} < 2021\\ \Leftrightarrow -\sqrt{\frac{2021}{2019}} < x< \sqrt{\frac{2021}{2019}} \ ( kết\ hợp\ ĐK)\\ \\ \Leftrightarrow 0\leqslant x< \sqrt{\frac{2021}{2019}}\\ Vậy\ S=( -\infty ;0) \cup \left[ 0;\sqrt{\frac{2021}{2019}}\right)\\ d.\ \ \sqrt{x^{2} +2021} < x\sqrt{2020}( 2)\\ TH1:\ x< 0\Rightarrow ( 2) \ vô\ lí\\ TH2:\ x\geqslant 0\\ ( 2) \Leftrightarrow x^{2} +2021< 2020x^{2}\\ \Leftrightarrow 2019x^{2} >2021\\ \Leftrightarrow x >\sqrt{\frac{2021}{2019}} \ \ or\ \ x< -\sqrt{\frac{2021}{2019}}\\ kết\ hợp\ ĐK\\ \Leftrightarrow x >\sqrt{\frac{2021}{2019}}\\ Vậy\ S=\left(\sqrt{\frac{2021}{2019}} ;+\infty \right) \end{array}$
