Question

Se considera ecuatia 2x²-4x-5=0. Folosind relatiile lui Viete, sa se calculeze valoarea expresiei $\frac{2x_1^2-3x_1-2}{2x_1^2+3x_1+2} + \frac{2x_2^2-3x_1-2}{2x_2^2+3x_2+2} .$
(eu cred ca la a 2-a fractie probabil trebuia la numarator 3x₂, dar asa scrie in carte, nu stiu cum e corect)

Answer 1

$2x^2-4x-5=0\\ \\ 2x_1^2-4x_1-5 = 0\Big|+7x_1+7 \Rightarrow 2x_1^2 +3x_1 + 2 = 7(x_1+1) \\ \\ 2x_2^2-4x_2-5 = 0 \Big| +7x_2+7 \Rightarrow 2x_2^2 +3x_2 + 2 = 7(x_2+1) \\ \\ 2x_1^2-4x_1-5 = 0\Big|+x_1+3 \Rightarrow 2x_1^2 -3x_1-2 = x_1+3 \\ \\ 2x_2^2-4x_2-5 = 0\Big|+x_2+3 \Rightarrow 2x_2^2 -3x_2-2 = x_2+3 \\ \\ \\ \Rightarrow \dfrac{2x_1^2 -3x_1-2}{2x_1^2 +3x_1 + 2} + \dfrac{2x_2^2 -3x_2-2}{2x_2^2 +3x_2 + 2} = \\ \\ = \dfrac{x_1+3}{7(x_1+1)} + \dfrac{x_2+3}{7(x_2+1)} = \\ \\ = \dfrac{1}{7} \cdot \Big(\dfrac{x_1+3}{x_1+1} + \dfrac{x_2+3}{x_2+1}\Big)$

De aici e usor, aducem la acelasi numitor paranteza si o sa iasa frumos.