Question

Calcule as somas das algébricas:

A) √75 + √12 - √48

B) √6 + 2√24 + √54 - √150

C) √32 - (-√50 + 4√8)

Answer 1

Oie, tudo bom?

a)

$= \sqrt{75} + \sqrt{12} - \sqrt{48} \\ = \sqrt{5 {}^{2} \: . \: 3 } + \sqrt{2 {}^{2} \: . \: 3} - \sqrt{4 {}^{2} \: . \: 3} \\ = \sqrt{5 {}^{2} } \sqrt{3} + \sqrt{2 {}^{2} } \sqrt{3} - \sqrt{4 {}^{2} } \sqrt{3} \\ = 5 \sqrt{3} + 2 \sqrt{3} - 4 \sqrt{3} \\ = (5 + 2 - 4) \sqrt{3} \\ = (7 - 4) \sqrt{3} \\ \boxed{ = 3 \sqrt{3} }$

b)

$= \sqrt{6} + 2 \sqrt{24} + \sqrt{54} - \sqrt{150} \\ = \sqrt{6} + 2 \sqrt{2 {}^{2} \: . \: 6 } + \sqrt{3 {}^{2} \: . \: 6 } - \sqrt{5 {}^{2} \: . \: 6} \\ = \sqrt{6} + 2 \sqrt{2 {}^{2} } \sqrt{6} + \sqrt{3 {}^{2} } \sqrt{6} - \sqrt{5 {}^{2} } \sqrt{6} \\ = \sqrt{6} + 2 \: . \: 2 \sqrt{6} + 3 \sqrt{6} - 5 \sqrt{6} \\ = \sqrt{6} + 4 \sqrt{6} + 3 \sqrt{6} - 5 \sqrt{6} \\ = 1 \sqrt{6} + 4 \sqrt{6} + 3 \sqrt{6} - 5 \sqrt{6} \\ = (1 + 4 + 3 - 5) \sqrt{6} \\ = (8 - 5) \sqrt{6} \\ \boxed{ = 3 \sqrt{6} }$

c)

$= \sqrt{32} - ( - \sqrt{50} + 4 \sqrt{8} ) \\ = \sqrt{2{}^{5} } - ( - \sqrt{5 {}^{2} \: . \: 2} + 4 \sqrt{2 {}^{3} } ) \\ = \sqrt{2 {}^{4 + 1} } - ( - \sqrt{5 {}^{2} } \sqrt{2} + 4 \sqrt{2 {}^{2 + 1} } ) \\ = \sqrt{2 {}^{4} \: . \: 2 {}^{1} } - ( - 5 \sqrt{2} + 4 \sqrt{2 {}^{2} \: . \: 2 {}^{1} } ) \\ = \sqrt{2 {}^{4} \: . \: 2 } - ( - 5 \sqrt{2} + 4 \sqrt{2 {}^{2} \: . \: 2} ) \\ = \sqrt{2{}^{4} } \sqrt{2} - ( - 5 \sqrt{2} + 4 \sqrt{2 {}^{2} } \sqrt{2} ) \\ = 2 {}^{2} \sqrt{2} - ( - 5 \sqrt{2} + 4 \: . \: 2 \sqrt{2} ) \\ = 2 {}^{2} \sqrt{2} - ( - 5 \sqrt{2} + 8 \sqrt{2} ) \\ = 4 \sqrt{2} - ( - 5 + 8) \sqrt{2} \\ = 4 \sqrt{2} - 3 \sqrt{2} \\ = (4 - 3) \sqrt{2} \\ = 1 \sqrt{2} \\ \boxed{ = \sqrt{2} }$

Att. NLE Top Shotta