Question

- DETERMINE AS SOMAS ALGEBRICAS:
$\frac{7}{3} \sqrt[3]{2} - 2 \sqrt[3]{2} - \frac{5}{4} \sqrt[3]{2}$
me ajudem pfffffffffff​

Answer 1

$~\huge\mid{\boxed{\bf{\blue{Matem\acute{a}tica}}}\mid}$

$\frac{7}{3} \sqrt[3]{2} - 2 \sqrt[3]{2} - \frac{5}{4} \sqrt[3]{2}$

$\frac{7}{3} \times \frac{ \sqrt[3]{2} }{1} - 2 \sqrt[3]{2} - \frac{5}{4} \times \frac{ \sqrt[3]{2} }{1}$

$\frac{7 \sqrt[3]{2} }{3} - 2 \sqrt[3]{2} - \frac{5 \sqrt[3]{2} }{4}$

$\frac{4 \times 7 \sqrt[3]{2} }{4 \times 3} - \frac{2 \sqrt[3]{2} }{1} - \frac{3 \times 5 \sqrt[3]{2} }{3 \times 4}$

$\frac{28 \sqrt[3]{2} }{12} - \frac{2 \sqrt[3]{2} }{1} - \frac{15 \sqrt[3]{2} }{12}$

$\frac{28 \sqrt[3]{2} }{12} - \frac{12 \times 2 \sqrt[3]{2} }{12} - \frac{15 \sqrt[3]{2} }{12}$

$\frac{28 \sqrt[3]{2} }{12} - \frac{24 \sqrt[3]{2} }{12} - \frac{15 \sqrt[3]{2} }{12}$

$\frac{28 \sqrt[3]{2} - 24 \sqrt[3]{2} - 15 \sqrt[3]{2} }{12}$

Resposta: $\color{green} \boxed{{  - \frac{11 \sqrt[3]{2} }{12} }}$

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