Vì \({5^2} = b\) nên \(\sqrt b  = a\). Hai số \(a,b\) thích hợp điền vào chỗ trống lần lượt là:

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

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

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

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

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

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

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

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

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

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

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

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

 \(1\)   

\(2\)    

 \(3\) 

\(4\)

\(121\) hoặc \( - 121\)

\(\sqrt {11} \) hoặc \( - \sqrt {11} \)

\(\dfrac{{11}}{2}\)

\(22\) hoặc \( - 22\)

\(3 \in Q\)

\(\dfrac{5}{3} \in Z\)

\(\dfrac{0}{3} \in Q\)

\(\dfrac{1}{2} \notin N\)