Răspuns :
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
= √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
[tex] 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[/tex]
