Question

2 si 4 va rog frumos!Dau coroana!!!​

Answer 1

Răspuns:

2.

a + 1/a = 3 |^2

a^2 + 2 + 1/a^2 = 9

=>a^2 + 1/a^2 = 7

4.

a=rad(4+rad12)=rad[4+rad(16-12)/2] + rad[4-rad(16-12)/2]

a=rad3 + 1

b=rad3 - 1

(rad3+1+rad3-1)^2=3

M.g.=rad[(rad3+1)(rad3-1)]=rad(3-1)=rad2

Answer 2

2)

${(a + \frac{1}{a}) }^{2} = {3}^{2} \\ {a}^{2} + 2 \times a \times \frac{1}{a} + \frac{1}{ {a}^{2} } = 9 \\ {a}^{2} + 2 + \frac{1}{ {a}^{2} } = 9 \\ {a}^{2} + \frac{1 }{ {a}^{2} } = 7$

4)

$a = \sqrt{4 + 2 \sqrt{3} } \\ 2 \times a \times b = 2 \sqrt{3} \\ a \times b = \sqrt{3} \\ a = \sqrt{3} \\ b = 1$

$\sqrt{4 + 2 \sqrt{3} } = | \sqrt{3} + 1| = \sqrt{3} + 1$

$b = \sqrt{4 - 2 \sqrt{3} } \\ 2 \times a \times b = 2 \sqrt{3} \\ a \times b = \sqrt{3} \\ a = \sqrt{3} \\ b = 1$

$\sqrt{4 - 2 \sqrt{3} } = | \sqrt{3} - 1 | = \sqrt{3} - 1$

a)

${(a + b)}^{2} = {( \sqrt{3} + 1 + \sqrt{3} - 1) }^{2} = {(2 \sqrt{3}) }^{2} = 12$

b)

$mg = \sqrt{ (\sqrt{3} + 1) \times ( \sqrt{3 } - 1)} \\ mg = \sqrt{ { (\sqrt{3} )}^{2} - 1} \\ mg = \sqrt{3 - 1} = \sqrt{2}$