Question

URGENT EX 2 dau coroana !!!!!!

Answer 1

a)

$E(x) = \bigg(\dfrac{x^{3} - 2x^{2}}{2x^{2}} - \dfrac{x^{2} - 4 }{4x} \bigg) : \dfrac{x-2}{2} + \dfrac{1}{2} = \bigg(\dfrac{x(x^{2} - 2x)}{2x^{2}} - \dfrac{x^{2} - 4 }{4x} \bigg) \cdot \dfrac{2}{x-2} + \dfrac{1}{2} =$

$= \bigg(\dfrac{x^{2} - 2x}{2x} - \dfrac{x^{2} - 4 }{4x} \bigg) \cdot \dfrac{2}{x - 2} + \dfrac{1}{2} = \dfrac{2x^{2} - 4x - x^{2} + 4}{4x} \cdot \dfrac{2}{x - 2} + \dfrac{1}{2}$

$= \dfrac{x^{2} - 4x + 4}{2x} \cdot \dfrac{1}{x - 2} + \dfrac{1}{2} = \dfrac{(x - 2)^{2}}{2x} \cdot \dfrac{1}{x - 2} + \dfrac{1}{2} = \dfrac{x - 2}{2x} + \dfrac{1}{2}$

$= \dfrac{x}{2x} - \dfrac{2}{2x} + \dfrac{1}{2} = \dfrac{1}{2} - \dfrac{1}{x} + \dfrac{1}{2} = \bf 1 - \dfrac{1}{x}$

b)

$E(x) = 1 - \dfrac{1}{x} = \dfrac{x - 1}{x}$

$E(3) \cdot E(4) \cdot E(5) \cdot E(6) \cdot E(7) \cdot E(8) = \dfrac{3 - 1}{3} \cdot \dfrac{4 - 1}{4} \cdot \dfrac{5 - 1}{5} \cdot \dfrac{6 - 1}{6} \cdot \dfrac{7 - 1}{7} \cdot \dfrac{8 - 1}{8} =$

$= \dfrac{2}{\not3} \cdot \dfrac{\not3}{\not4} \cdot \dfrac{\not4}{\not5} \cdot \dfrac{\not5}{\not6} \cdot \dfrac{\not6}{\not7} \cdot \dfrac{\not7}{8} = \dfrac{2}{8} = \bf \dfrac{1}{4}$