Bất phương trình căn bậc hai của (x^2+6x-5)>8-2x có nghiệm

Ta có\[\sqrt { - {x^2} + 6x - 5} >8 - 2x\]

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{ - {x^2} + 6x - 5 \ge 0}\\{8 - 2x < 0}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{8 - 2x \ge 0}\\{ - {x^2} + 6x - 5 >{{(8 - 2x)}^2}}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{1 \le x \le 5}\\{x >4}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \le 4}\\{ - 5{x^2} + 38x - 69 >0}\end{array}} \right.}\end{array}} \right.\)</>

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{1 \le x \le 5}\\{x >4}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \le 4}\\{3 < x < \frac{{23}}{5}}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{4 < x \le 5}\\{3 < x \le 4}\end{array}} \right. \Leftrightarrow 3 < x \le 5\)