P=1/6 Q=2/3
Fazemos a resolução:
[tex]\boxed{\begin{array}{lr} \large \sf 2\times\left(\dfrac{1}{6}\right)^{2}-4\times\dfrac{1}{\diagup\!\!\!\!6}\times\dfrac{\diagup\!\!\!\!2}{3}-3\times\left(\dfrac{2}{3}\right)^{2} \\ \\ \large \sf \diagup\!\!\!\!2\times\dfrac{1}{\diagup\!\!\!\!\!36}-4\times\dfrac{1}{3}\times\dfrac{1}{3}-\diagup\!\!\!\!3\times\dfrac{4}{\diagup\!\!\!\!9}\\ \\ \large \sf \dfrac{1}{18}-\dfrac{4}{9}-\dfrac{4}{3}\to mmc\:18{,}9{,}3=18 \\ \\ \large \sf \dfrac{1-8-24}{18} \\ \\ \large \sf \dfrac{-31}{18} \end{array}}[/tex]
- Logo fazemos a formula para inverter o sinal negativo:
[tex]\large\boxed{\boxed{\sf \dfrac{-a}{b}=\dfrac{a}{-b}=-\dfrac{a}{b}}}[/tex]
Aplicamos a formula:
[tex]\boxed{\boxed{\sf \dfrac{-31}{18}=\dfrac{31}{-18}=-\dfrac{31}{18}}}[/tex]
Logo a resposta será:
[tex]\boxed{\boxed{\sf -\dfrac{31}{18}}}\Longleftarrow[/tex]
Bons estudos =)
