Duas frações são equivalentes quando têm a mesma forma irredutível. Primeiro achamos o nome mais limpo de $\mathbf{\frac{4}{6}}$ , cortando o numerador e o denominador pelo divisor comum 2:

$\mathbf{\dfrac{\textcolor{#ef4444}{4}}{\textcolor{#3b82f6}{6}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{4} \div 2}{\textcolor{#3b82f6}{6} \div 2}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{2}}{\textcolor{#3b82f6}{3}}}$

Agora levamos a opção B, $\mathbf{\frac{8}{12}}$ , ao mesmo teste, cortando pelo divisor comum 4:

$\mathbf{\dfrac{\textcolor{#ef4444}{8}}{\textcolor{#3b82f6}{12}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{8} \div 4}{\textcolor{#3b82f6}{12} \div 4}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{2}}{\textcolor{#3b82f6}{3}}}$

As duas chegam a $\mathbf{\frac{2}{3}}$ , então são equivalentes. As outras dão nomes diferentes: $\mathbf{\frac{\textcolor{#ef4444}{6}}{\textcolor{#3b82f6}{8}}}$ e $\mathbf{\frac{\textcolor{#ef4444}{12}}{\textcolor{#3b82f6}{16}}}$ valem $\mathbf{\frac{\textcolor{#ef4444}{3}}{\textcolor{#3b82f6}{4}}}$ ; $\mathbf{\frac{\textcolor{#ef4444}{2}}{\textcolor{#3b82f6}{4}}}$ vale $\mathbf{\frac{\textcolor{#ef4444}{1}}{\textcolor{#3b82f6}{2}}}$ ; e $\mathbf{\frac{\textcolor{#ef4444}{8}}{\textcolor{#3b82f6}{10}}}$ vale $\mathbf{\frac{\textcolor{#ef4444}{4}}{\textcolor{#3b82f6}{5}}}$ .

O gabarito é letra B.

Questão Q14800