Question

urgeeeent, dau coronițăăă

Answer 1

$a) \sqrt{1 + 3 + 5 + ... + 21}$

$1 + 3 + 5 + ... + 2n - 1 = {n}^{2}$

$2n - 1 = 21$

$2n = 21 + 1$

$2n = 22 \: | \div 2$

$n = 11$

$\sqrt{ {n}^{2} } = \sqrt{ {11}^{2} } = \sqrt{121} = 11 \: \in \: \mathbb{N}$

$b) \sqrt[3]{27} - \sqrt{12} + 2 \sqrt{3}$

$= \sqrt[3]{ {3}^{3} } - 2 \sqrt{3} + 2 \sqrt{3}$

$= {3}^{ \frac{3}{3} } + 0$

$= {3}^{1}$

$= 3 \: \in \: \mathbb{Z}$

$c) {(1 + \sqrt{2} )}^{2} + {(1 - \sqrt{2} )}^{2}$

$= {1}^{2} + 2 \times 1 \times \sqrt{2} + ({ \sqrt{2} )}^{2} +{1}^{2}- 2 \times 1 \times \sqrt{2} + { (\sqrt{2}) }^{2}$

$= 1 + 2 \sqrt{2} + 2 + 1 - 2 \sqrt{2} + 2$

$= 1 + 1 + 2 + 2 + 2 \sqrt{2} - 2 \sqrt{2}$

$= 6 + 0$

$= 6 \: \in \: \mathbb{N}$