[tex]\it a = 2^{1990}-2^{1989}-2^{1988}=2^{1988}(2^2-2-1) =2^{1988}(4-2-1) =2^{1988}\\ \\ \dfrac{4^{993}}{0,25}= \dfrac{(2^2)^{993}}{ \dfrac{1}{4}}= \dfrac{2^{1986}}{ \dfrac{1}{2^2}}= \dfrac{2^{1988}}{1}[/tex]
Relația din enunț devine:
[tex]\it \dfrac{2^{1998}}{x}= \dfrac{2^{1988}}{1} \Rightarrow x=1[/tex]
