[tex]\displaystyle \left \{ {{x^2-5x+7\geq 1} \atop {-x^2+17x-16>0}} \right. =\left \{ {{x \in (- \infty,2] \cup[3, + \infty)} \atop {x \in (1,16)}} \right. \\\\x \in (1,2]\cup[3,16)[/tex]
