Nếu ${x^2} = 11$ thì $x$ bằng:

$\dfrac{5}{{15}} = \dfrac{{ - 9}}{{ - 27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{{ - 9}};\,\dfrac{5}{{ - 9}} = \dfrac{{15}}{{ - 27}};\,\dfrac{{ - 9}}{5} = \dfrac{{ - 27}}{{15}}$

$\dfrac{5}{{15}} = \dfrac{{ - 9}}{{ - 27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{{ - 9}};\,\dfrac{5}{{ - 9}} = \dfrac{{15}}{{ - 27}};\,\dfrac{{ - 9}}{5} = \dfrac{{ - 15}}{{27}}$

$\dfrac{5}{{15}} = \dfrac{{ - 9}}{{ - 27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{9}$

$\dfrac{{15}}{5} = \dfrac{9}{{27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{{ - 9}};\,\dfrac{5}{{ - 9}} = \dfrac{{15}}{{ - 27}};\,\dfrac{{ - 9}}{5} = \dfrac{{ - 15}}{{27}}$

${\left( {-2019} \right)^0} = 1$

$\left( {0,5} \right).{\left( {0,5} \right)^2} = \dfrac{1}{4}$

${4^6}:{\rm{ }}{4^4} = 16$

${\left( {-3} \right)^3}.{\left( {-{\rm{ }}3} \right)^{{\rm{ }}2}} = {\left( { - 3} \right)^5}$

$\widehat {zOt} = \widehat {z'Ot'} = 72^\circ$

$\widehat {zOt} = \widehat {z'Ot'} = 30^\circ$

$\widehat {zOt} = \widehat {z'Ot'} = 36^\circ$

$\widehat {zOt} = 72^\circ ;\,\widehat {z'Ot'} = 36^\circ$

$a = 25$ và $b = 25$

$a = 5$ và $b = 25$

$a = 5$ và $b = 10$

$a = 5$ và $b = 5$

 $1$   

$2$    

 $3$ 

$4$

$3 \in Q$

$\dfrac{5}{3} \in Z$

$\dfrac{0}{3} \in Q$

$\dfrac{1}{2} \notin N$