Question

sa se calculeze E1=1/x1^3+1/x2^3 si E2=|x1-x2|​

Answer 1

$E_1 = \frac{1}{x_1^3} + \frac{1}{x_2^3} = \frac{x_1^3 + x_2^3}{x_1^3\cdot x_2^3}$

$E_2 = |x_1 - x_2| = \begin{cases}x_1 - x_2, \quad \text{daca }x_1 - x_2 \geq 0 \Rightarrow x_1 \geq x_2\\x_2 - x_1, \quad \text{daca } x_1 - x_2 < 0 \Rightarrow x_1 < x_2\end{cases}\\\\E_2 = \begin{cases}x_1 - x_2, \quad \text{daca }x_1 \geq x_2\\x_2 - x_1, \quad \text{daca } x_1 < x_2\end{cases}$

$x^2 - x - a^2 = 0 \\\\\Delta = 1 + 4a^2 \Rightarrow \sqrt{\Delta} = \sqrt{4a^2 + 1}\\\\x_{1,2} = \frac{1 \pm \sqrt{4a^2 + 1}}{2} \Rightarrow E_2 = \begin{cases}x_1 - x_2, \quad \text{daca }x_1 \geq x_2\\x_2 - x_1, \quad \text{daca } x_1 < x_2\end{cases}\\\\x_1 < x_2 \\\\\Rightarrow E_2 = x_2 - x_1 = \frac{1+\sqrt{4a^2 + 1}}{2} - \frac{1-\sqrt{4a^2+1}}{2} = \sqrt{4a^2 + 1}\\\\S = \frac{-b}{a} = \frac{1}{1} = 1\\\\P = \frac{c}{a} = \frac{-a^2}{1} = -a^2\\\\E_1 = \frac{S^3 - 3PS}{P^3} = \frac{1^3 - 3\cdot (-a^2) \cdot 1}{(-a^2)^3} = \frac{1+3a^2}{-a^6}$