Question

$a = \sqrt{27} - 4 \sqrt{3} + \sqrt{12}$
$b = \sqrt{18} - \sqrt{32} + \sqrt{18}$
Comparați a și b!

Answer 1

a = √27 - 4√3 + √12

= √3²·3 - 4√3 + √2²·3

= 3√3 - 4√3 + 2√3

= √3


b = √18 - √32 + √18

= √2·3² - √2·4² + √2·3²

= 3√2 - 4√2 + 3√2

= 6√2 - 4√2

= 2√2

= √8

a < b

Answer 2

$a = \sqrt{27} \: - \: \it 4 \sqrt{3} \: + \: \sqrt{12} \\ \it a = 3 \sqrt{3 } \: - \: 4 \sqrt{3} \: + \: 2 \sqrt{3} \\ \it a = - \: \sqrt{3} \: + \: 2 \sqrt{3} \\ \it a = \sqrt{3} \\ \it b = \sqrt{18} \: - \: \sqrt{32} \: + \: \sqrt{18} \\ \it b = 3 \sqrt{2 } \: - \: 4 \sqrt{2 } \: + \: 3 \sqrt{2} \\ \it b = - \: \sqrt{2} \: + \: 3 \sqrt{2} \\ \it b = 2 \sqrt{2} = \sqrt{ {2}^{2} \cdot 2 } = \sqrt{4 \cdot 2} = \sqrt{8} \\ \it \sqrt{3} < \sqrt{8} \: < = > \: a < b$