Question

am si eu nevoie sa imi rezolvati ex 16 cu formulele de la ex 15​

Answer 1

Explicație pas cu pas:

a)

$a = 5; \ b = 24 \\ {c}^{2} = {a}^{2} - b = {5}^{2} - 24 = 25 - 24 = 1 \\ \implies c = 1$

$\sqrt{5 + \sqrt{24}} = \sqrt{ \frac{5 + 1}{2} } + \sqrt{ \frac{5 - 1}{2} } = \sqrt{ \frac{6}{2} } + \sqrt{ \frac{4}{2} } = \sqrt{3} + \sqrt{2}$

$\sqrt{5 - \sqrt{24}} = \sqrt{ \frac{5 + 1}{2} } - \sqrt{ \frac{5 - 1}{2} } = \sqrt{ \frac{6}{2} } - \sqrt{ \frac{4}{2} } = \sqrt{3} - \sqrt{2}$

$\sqrt{5 + \sqrt{24}} + \sqrt{5 - \sqrt{24}} = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{R-Q}$

b)

$a = 9; \ b = 80 \\ {c}^{2} = {a}^{2} - b = 81 - 80 = 1 \\ \implies c = 1$

$\sqrt{9 - 4\sqrt{5}} = \sqrt{9 - \sqrt{80}} = \sqrt{ \frac{9 + 1}{2} } - \sqrt{ \frac{9 - 1}{2} } = \sqrt{ \frac{10}{2} } - \sqrt{ \frac{8}{2} } = \sqrt{5} - \sqrt{4} = \sqrt{5} - 2$

$\sqrt{9 + 4\sqrt{5}} = \sqrt{9 + \sqrt{80}} = \sqrt{ \frac{9 + 1}{2} } + \sqrt{ \frac{9 - 1}{2} } = \sqrt{ \frac{10}{2} } + \sqrt{ \frac{8}{2} } = \sqrt{5} + \sqrt{4} = \sqrt{5} + 2$

$\sqrt{9 - 4\sqrt{5}} + \sqrt{9 + 4\sqrt{5}} = \sqrt{5} - 2 + \sqrt{5} + 2 = 2 \sqrt{5} \in \mathbb{R-Q}$

c)

$a = 7; \ b = 48 \\ {c}^{2} = {a}^{2} - b = 49 - 48 = 1 \\ \implies c = 1$

$\sqrt{7 + 4\sqrt{3}} = \sqrt{7 + \sqrt{48}} = \sqrt{ \frac{7 + 1}{2} } + \sqrt{ \frac{7 - 1}{2} } = \sqrt{ \frac{8}{2} } + \sqrt{ \frac{6}{2} } = \sqrt{4} + \sqrt{3} = 2 + \sqrt{3}$

$\sqrt{7 - 4\sqrt{3}} = \sqrt{7 - \sqrt{48}} = \sqrt{ \frac{7 + 1}{2} } - \sqrt{ \frac{7 - 1}{2} } = \sqrt{ \frac{8}{2} } - \sqrt{ \frac{6}{2} } = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3}$

$\sqrt{7 + 4\sqrt{3}} + \sqrt{7 + 4\sqrt{3}} = 2 + \sqrt{3} + 2 - \sqrt{3} = 4 \in \mathbb{Q}$

d)

$a = 17; \ b = 288 \\ {c}^{2} = {a}^{2} - b = 289 - 288 = 1 \\ \implies c = 1$

$\sqrt{17 + 12\sqrt{2}} = \sqrt{17 + \sqrt{288}} = \sqrt{ \frac{17 + 1}{2} } + \sqrt{ \frac{17 - 1}{2} } = \sqrt{ \frac{18}{2} } + \sqrt{ \frac{16}{2} } = \sqrt{9} + \sqrt{8} = 3 + 2\sqrt{2}$

$\sqrt{17 - 12\sqrt{2}} = \sqrt{17 - \sqrt{288}} = \sqrt{ \frac{17 + 1}{2} } - \sqrt{ \frac{17 - 1}{2} } = \sqrt{ \frac{18}{2} } - \sqrt{ \frac{16}{2} } = \sqrt{9} - \sqrt{8} = 3 - 2\sqrt{2}$

$\sqrt{17 + 12\sqrt{2}} + \sqrt{17 - 12\sqrt{2}} = 3 + 2\sqrt{2} + 3 - 2 \sqrt{2} = 6 \in \mathbb{Q}$