$- 1 \le 2{x^2} - 3 < 2$ or $2 \le 2{x^2} < 5$ or $1 \le {x^2} < {5 \over 2}$ $x \in \left( { - \sqrt {{5 \over 2}} , - 1} \right] \cup \left[ {1,\sqrt {{5 \over 2}} } \right)$ ${\log _{{1 \over 2}}}({x^2} - 5x + 5) > 0$ $0 < {x^2} - 5x + 5 < 1$ ${x^2} - 5x + 5 > 0$ & ${x^2} - 5x + 4 < 0$ $x \in \left( { - \infty ,{{5 - \sqrt 5 } \over 2}} \right) \cup \left( {{{5 + \sqrt 5 } \over 2},\infty } \right)$ & $x \in ( - \infty ,1) \cup (4,\infty )$ Taking intersection $x \in \left( {1,{{5 - \sqrt 5 } \over 2}} \right)$