Question

essas expressões são quais

Answer 1

Olá, tudo bem? Vamos resolver uma a uma:

a) $\left(\dfrac{3}{2}\right)^2-\left(\dfrac{5}{6}\right)=\dfrac{9}{4}-\dfrac{5}{6}=\dfrac{27-10}{12}=\boxed{\dfrac{17}{12}}$

b) $3^2-\dfrac{7}{6}+\left(\dfrac{3}{4}\right)^0=9-\dfrac{7}{6}+1=10-\dfrac{7}{6}=\boxed{\dfrac{53}{6}}$

c) $\left[\dfrac{8}{3}+\left(\dfrac{1}{2}\right)^2\right]\div\left[\dfrac{3}{4}-\left(\dfrac{1}{2}\right)^1\right]=\left(\dfrac{8}{3}+\dfrac{1}{4}\right)\div \left(\dfrac{3}{4}-\dfrac{1}{2}\right)=$

$=\left(\dfrac{32+3}{12}\right)\div\left(\dfrac{3-2}{4}\right)=\dfrac{\frac{35}{12}}{\frac{1}{4}}=\dfrac{140}{12}=\boxed{\dfrac{35}{3}}$

d) $2\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^2+\dfrac{1}{5}\div\dfrac{3}{5}+\left(\dfrac{2}{3}\right)^0=\dfrac{7}{3}-\dfrac{1}{4}+\dfrac{1}{3}+1=$

$\dfrac{28-3+4+12}{12}=\boxed{\dfrac{41}{12}}$

e) $\left(\dfrac{1}{2}\right)^3+\left[\dfrac{2}{5}+\dfrac{3}{5}-\left(\dfrac{7}{8}\right)^0\right]=\dfrac{1}{8}+0=\boxed{\dfrac{1}{8}}$

É isso!! :)