iрраціональні нерівності

$$\sqrt[2k]{f(x)}>\sqrt[2k]{g(x)}\iff \begin{array}{c}\{\begin{array}{l}f(x)>g(x)\\ g(x)\ge 0\end{array}\end{array}$$

$$\sqrt[2k]{f(x)}<g(x)\iff \begin{array}{c}\{\begin{array}{l}g(x)>0\\ f(x)<(g(x){)}^{2k}\\ f(x)\ge 0\end{array}\end{array}$$

$$\sqrt[2k]{f(x)}\le g(x)\iff \begin{array}{c}\{\begin{array}{l}g(x)\ge 0\\ f(x)\le (g(x){)}^{2k}\\ f(x)\ge 0\end{array}\end{array}$$

$$\sqrt[2k]{f(x)}>g(x)\iff [\begin{array}{c}\begin{array}{c}\{\begin{array}{l}g(x)\ge 0\\ f(x)>(g(x){)}^{2k}\end{array}\end{array}\\ \begin{array}{c}\{\begin{array}{l}g(x)<0\\ f(x)\ge 0\end{array}\end{array}\end{array}$$

$$\sqrt[2k]{f(x)}\ge g(x)\iff [\begin{array}{c}\begin{array}{c}\{\begin{array}{l}g(x)\ge 0\\ f(x)\ge (g(x){)}^{2k}\end{array}\end{array}\\ \begin{array}{c}\{\begin{array}{l}g(x)<0\\ f(x)\ge 0\end{array}\end{array}\end{array}$$

$$f(x)\sqrt[2k]{g(x)}\ge 0\iff [\begin{array}{c}\begin{array}{c}\{\begin{array}{l}g(x)\ge 0\\ f(x)\ge (g(x){)}^{2k}\end{array}\end{array}\\ \begin{array}{c}\{\begin{array}{l}g(x)<0\\ f(x)\ge 0\end{array}\end{array}\end{array}$$