Question

ex 5 va rog, dau coroana

Answer 1

Răspuns:

$a)x + \frac{1}{x} = 4 \\ {(x + \frac{1}{x}) }^{2} = {x}^{2} + 2 \times x \times \frac{1}{x} + \frac{1}{ {x}^{2} } = {x}^{2} + 2 + \frac{1}{ {x}^{2} } \\ {x}^{2} + \frac{1}{ {x}^{2} } = {(x + \frac{1}{x} })^{2} - 2 = {4}^{2} - 2 = 16 - 2 = 14$

$b)x - \frac{1}{x} = 2 \\ {(x - \frac{1}{x} })^{2} = {x}^{2} - 2 \times x \times \frac{1}{x} + \frac{1}{ {x}^{2} } = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ {x}^{2} + \frac{1}{ {x}^{2} } = ( {x - \frac{1}{x} })^{2} + 2 = {2}^{2} + 2 = 6 \\ {x}^{2} + \frac{1}{ {x}^{2} } - 20 = 6 - 20 = - 14$

$c) {(x + \frac{1}{x} })^{2} = {x}^{2} + 2 \times x \times \frac{1}{x} + \frac{1}{ {x}^{2} } = {x}^{2} + 2 + \frac{1}{ {x}^{2} } \\ {(x + \frac{1}{x} })^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 5 + 2 = 7 \\ x + \frac{1}{ {x} } = \sqrt{7}$

$d)( {x + \frac{1}{x} })^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\ ({x + \frac{1}{x} })^{2} = 7 + 2 = 9 \\ x + \frac{1}{x} = \sqrt{9} = 3$

$e) {(x - \frac{1}{x} })^{2} = {x}^{2} - 2 \times x \times \frac{1}{x} + \frac{1}{ {x}^{2} } = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ ( {x - \frac{1}{x} })^{2} = 5 - 2 = 3 \\ x - \frac{1}{x} = \sqrt{3} \\ |x - \frac{1}{x} | = | \sqrt{3} | = \sqrt{3}$