Question

Simplificando os radicais, 2√3 + 2√12 - 2√75, obtém-se:

a) -2√3
b) 12√3
c) -4√3
d) 10√3

Answer 1

$\Large\boxed{\begin{array}{l}\begin{array}{c|c}\sf12&\sf2\\\sf6&\sf2\\\sf3&\sf3\\\sf1\end{array}\\\sf 12=2^2\cdot3\\\sf \sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}\\\begin{array}{c|c}\sf75&\sf3\\\sf25&\sf5\\\sf5&\sf5\\\sf1\end{array}\\\sf\sqrt{75}=\sqrt{3\cdot5^2}=5\sqrt{3}\\\sf2\sqrt{3}+2\sqrt{12}-2\sqrt{75}=2\sqrt{3}+2\cdot2\sqrt{3}-2\cdot5\sqrt{3}\\\sf=2\sqrt{3}+4\sqrt{3}-10\sqrt{3}=-4\sqrt{3}\checkmark\end{array}}$

Answer 2

Resposta:

C) -4V3

$2 \sqrt{3} + 2 \sqrt{12} - 2 \sqrt{75} = \\ 2 \sqrt{3} + 2 \sqrt{12} - 2 \sqrt{75} \\ 2 \sqrt{12} \\ 2 \sqrt{4 \times 3} \\ 2 \sqrt{2 {}^{2} \times 3 } \\ 2 \sqrt{2 {}^{2}} \sqrt{3} \\2 \times 2 \sqrt{3} \\ 4 \sqrt{3} \\ 2 \sqrt{3} + 2 \sqrt{12} - 2 \sqrt{75} \\ 2 \sqrt{3} + 4 \sqrt{3} - 2 \sqrt{75} \\ 2 \sqrt{3} + 2 \sqrt{12} - 2 \sqrt{75} \\ - 2 \sqrt{75} \\ - 2 \sqrt{25 \times 3} \\ - 2 \sqrt{5 {}^{2} \times 3} \\ - 2 \sqrt{5 {}^{2} \sqrt{3}} \\ - 2 \times 5 \sqrt{3} \\ - 10 \sqrt{3} \\ 2 \sqrt{3} +2 \sqrt{12} - 2 \sqrt{75} \\ 2 \sqrt{3} + 4 \sqrt{3} - 10 \sqrt{3} \\ (2 + 4 - 10) \sqrt{3} \\ - 4 \sqrt{3} \\ 2 \sqrt{3} + 4 \sqrt{3} - 10 \sqrt{3} \\ - 4 \sqrt{3} \\ resposta- 4 \sqrt{3}$