Question

Efetue: Ajuda por favor!

a) √3 + √27
b) √75 - √12
c) 7√2 + √50
d) √12 - √75 + √3
e) 3√20 + √32 - 2√45 + √50
f) √125 + 2√27 - √20 + 3√12

Answer 1

$\sqrt{3} + \sqrt{27} = \sqrt{3} + \sqrt{ 3^{3} } = \sqrt{3} +3 \sqrt{3} = 4 \sqrt{3} \\ \\ \sqrt{75} - \sqrt{12} = \sqrt{3. 5^{2} }- \sqrt{3. 2^{2} } = 5 \sqrt{3} -2 \sqrt{3} = 3 \sqrt{3} \\ \\ 7 \sqrt{2} + \sqrt{50}= 7 \sqrt{2} + \sqrt{2. 5^{2} } = 7 \sqrt{2} +5 \sqrt{2} = 12 \sqrt{2} \\ \\ \sqrt{12} - \sqrt{75} + \sqrt{3} = 2 \sqrt{3} -5 \sqrt{3} + \sqrt{3} = 3 \sqrt{3} -5 \sqrt{3} = -2 \sqrt{3} \\ \\ 3\sqrt{20} + \sqrt{32} -2 \sqrt{45} + \sqrt{50} =$

$= 3 \sqrt{ 2^{2}.5} + \sqrt{2^5} -2 \sqrt{3^2.5} + \sqrt{2.5^2} = 6 \sqrt{5} +4 \sqrt{2} -6 \sqrt{5} +5 \sqrt{2} = 9 \sqrt{2} \\ \\ \sqrt{125} +2 \sqrt{27} - \sqrt{20} +3 \sqrt{12} = \sqrt{5^3} +6 \sqrt{3} -2 \sqrt{5} +6 \sqrt{3} = \\ \\ 5 \sqrt{5} -2 \sqrt{5} +12 \sqrt{3} = 3 \sqrt{5} +12 \sqrt{3}$