Każdą z liczb a i b zapiszmy w jak najprostszej postaci. 

$a=\frac{\sqrt[3]{9^{\frac{3}{2}}}+{\left(1\frac{7}{9}\right)}^{\frac{1}{2}}}{{\left(\sqrt{1}\right)}^{\frac{1}{2}}+{\left(1\frac{1}{2}\right)}^{-2}}=\frac{9^{\frac{{\sout{3}}^1}{2}\cdot \frac{1}{{\sout{3}}_1}}+{\left(\frac{16}{9}\right)}^{\frac{1}{2}}}{1+{\left(\frac{3}{2}\right)}^{-2}}=$  

$=\frac{9^{\frac{1}{2}}+\sqrt{\frac{16}{9}}}{1+{\left(\frac{2}{3}\right)}^2}=\frac{\sqrt{9}+\frac{4}{3}}{1+\frac{4}{9}}=\frac{3+1\frac{1}{3}}{1\frac{4}{9}}=$  

$=\frac{4\frac{1}{3}}{1\frac{4}{9}}=\frac{\frac{13}{3}}{\frac{13}{9}}=\frac{{\sout{13}}^1}{3}\cdot \frac{9}{{\sout{13}}_1}=\frac{1}{{\sout{3}}_1}\cdot {\sout{9}}^3=3$  

 

$b=\frac{\sqrt{\frac{1}{9}}\cdot \sqrt{2^2\cdot 5+2^{-2}}}{{\left(\sqrt{16}\right)}^{-\frac{1}{2}}}=\frac{\frac{1}{3}\cdot \sqrt{4\cdot 5+{\left(\frac{1}{2}\right)}^2}}{4^{-\frac{1}{2}}}=$    

$=\frac{\frac{1}{3}\cdot \sqrt{20+\frac{1}{4}}}{{\left(\frac{1}{4}\right)}^{\frac{1}{2}}}=\frac{\frac{1}{3}+\sqrt{20\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=\frac{\frac{1}{3}+\sqrt{\frac{81}{4}}}{\frac{1}{2}}=$  

$=\frac{\frac{1}{3}+\frac{9}{2}}{\frac{1}{2}}=\frac{\frac{2}{6}+\frac{27}{6}}{\frac{1}{2}}=\frac{\frac{29}{6}}{\frac{1}{2}}=\frac{29}{{\sout{6}}_3}\cdot {\sout{2}}^1=\frac{29}{8}=3\frac{5}{8}$  

 

Porównujemy teraz liczby a i b. 

$3<3\frac{5}{8}$  

więc: 

$\mathbf{a}<\mathbf{b}$