[tex]3-[-\frac12-(0,1+\frac14)]=3-[-\frac12-(\frac1{10}+\frac14)]=\frac{60}{20}-[\frac{-10-2-5}{20}]=[/tex]
[tex]=\frac{60+17}{20}=\frac{77}{20}=3,85[/tex]
Observe que:
[tex]0,1=\dfrac{1}{10}[/tex]
Seja [tex]\text{X}=3-\left\lceil-\dfrac{1}{2}-\left(\dfrac{1}{10}+\dfrac{1}{4}\right)\right\rceil[/tex]
Desta maneira, temos que:
[tex]\text{X}=3-\left\lceil-\dfrac{1}{2}-\dfrac{7}{20}\right\rceil[/tex]
Note que [tex]\text{mmc}(2, 20) = 20[/tex]
[tex]\text{X}=3+\dfrac{17}{20}[/tex]
Observe que [tex]\text{mmc}(1, 20) = 20[/tex].
Contudo, podemos afirmar que:
[tex]\text{X}=\dfrac{60+17}{20}=\dfrac{77}{20}[/tex]
