[tex]\it 3\dfrac{1}{2}=\dfrac{7}{2}\\ \\ \sqrt{0,0225}=0,15=\dfrac{\ 15^{(5}}{100}=\dfrac{3}{20}\\ \\ 0,(6)=\dfrac{\ 6^{(3}}{9}=\dfrac{2}{3}\\ \\ \sqrt{1,125\cdot2}=\sqrt{2,25}=1,5=\dfrac{15^{(5}}{10}=\dfrac{3}{2}\\ \\ 0,2(7)=\dfrac{27-2}{90}=\dfrac{\ 25^{(5}}{90}=\dfrac{5}{18}\\ \\ \sqrt{0,09}=0,3=\dfrac{3}{10}[/tex]
Acum, exercițiul devine:
[tex]\it \dfrac{7}{2}\cdot\dfrac{3}{20}-\dfrac{2}{3}\cdot \dfrac{3}{2}+\dfrac{5}{18}\cdot\dfrac{3}{10}=\dfrac{^{3)}21}{40}\ -\ ^{120)}1+\dfrac{^{10)}1}{12}= \dfrac{63-120+10}{120}=-\dfrac{47}{120}[/tex]
