[tex]\it 0,6(3)+\dfrac{^{100)}0,25}{\ \ 0,75}-0,(3)=\dfrac{0,6(3)}{a} \Rightarrow \dfrac{63-6}{90}+\dfrac{\ \ 25^{(25}}{75}-\dfrac{\ \ 3^{(3}}{9}=\dfrac{\dfrac{63-6}{90}}{a} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{57}{90}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{57}{90a} \Rightarrow \dfrac{57}{90}=\dfrac{57}{90a}\Big |_{\cdot\dfrac{90}{57}} \Rightarrow 1=\dfrac{1}{a} \Rightarrow a=1[/tex]
