A diferença é uma subtração: $\mathbf{\frac{3}{4}}$ − $\mathbf{\frac{2}{3}}$ . Os denominadores 4 e 3 são diferentes e nenhum é múltiplo do outro, então o denominador comum é 4 × 3 = 12.

Reescrevo o $\mathbf{\frac{\textcolor{#ef4444}{3}}{\textcolor{#3b82f6}{4}}}$ em doze avos:

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

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

= $\mathbf{\dfrac{\textcolor{#ef4444}{9}}{\textcolor{#3b82f6}{12}}}$

Reescrevo o $\mathbf{\frac{\textcolor{#ef4444}{2}}{\textcolor{#3b82f6}{3}}}$ em doze avos:

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

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

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

Agora subtraio os numeradores e conservo o denominador:

$\mathbf{\dfrac{\textcolor{#ef4444}{9}}{\textcolor{#3b82f6}{12}} - \dfrac{\textcolor{#ef4444}{8}}{\textcolor{#3b82f6}{12}}}$

= $\mathbf{\dfrac{\textcolor{#ef4444}{1}}{\textcolor{#3b82f6}{12}}}$

A alternativa $\mathbf{1}$ vem de subtrair os de cima e os de baixo separadamente, e a $\mathbf{\frac{17}{12}}$ vem de somar em vez de subtrair.

O gabarito é letra C.

Questão Q12388