log 3 ( 𝑥 − 3 ) < 1 ${\displaystyle {\log }_3\left(x-3\right)<1^{}}$

5 − 𝑥 + 2 > 0,2 ${\displaystyle 5^{-x+2}>\mathrm{0,2}}$

𝑥 − 3 ( 𝑥 − 6 ) 2 > 0 ${\displaystyle \frac{{}^{}x-3^{}}{{\left(x-6\right)}^2}>0}$

( 𝑥 − 6 ) ( 𝑥 − 3 ) > 0 ${\displaystyle \left(x-6\right)\left(x-3\right)>0^{}}$

3 < 𝑥 < 6 $3<x<6_{}^{}$ или 𝑥 > 6 $x>6_{}^{}$

3 < 𝑥 < 6 $3<x<6_{}^{}$

𝑥 < 3 $x<3_{}^{}$

𝑥 < 3 $x<3_{}^{}$ или 𝑥 > 6 $x>6_{}^{}$