Question

CINE MA AJUTA SI PE MINE LA ACEST EXERCITIU ?? :o3
1) Se considera expresia:
E(x)= x-3supra x-5 : x²-x-6supra x²-4x-5 +1suprax+2) ·5x+7supra3x+4 , x∈R {-4supra5; -2; -1; 3;5}
a) Aduceti expresia la forma cea mai simpla.
b) Aratati ca pentru orice n∈N fractia 5n+7supra 3n+4 este ireductubila
RASPUNS: a) E(x)= 5x+7supra3x+4
AJUTATI-MA VA ROG :o3

Answer 1

$E(x)= \frac{x-3}{x-5} : \frac{ x^{2} -x-6}{ x^{2} -4x-5} + \frac{1}{x+2} = \\ = \frac{x-3}{x-5} * \frac{ x^{2} -4x-5}{ x^{2} -x-6} + \frac{1}{x+2} = \\ = \frac{x-3}{x-5} * \frac{ x^{2} -1-4x-4}{ x^{2} -4-x-2} + \frac{1}{x+2} = \\ = \frac{x-3}{x-5} * \frac{(x+1)(x-1)-4(x+1)}{(x+2)(x-2)-(x+2)} + \frac{1}{x+2} = \\ = \frac{x-3}{x-5} * \frac{(x+1)(x-1-4)}{(x+2)(x-2-1)} + \frac{1}{x+2} = \\ = \frac{x-3}{x-5} * \frac{(x+1)(x-5)}{(x+2)(x-3)} + \frac{1}{x+2} = \\ = \frac{x+1}{x+2} + \frac{1}{x+2} =$
$= \frac{x+2}{x+2}= \\ =1.$

b) Fie d∈N astfel incat d | 5n+7 si d | 3n+4.
d | 5n+7 => d | 15n+21
d | 3n+4 => d | 15n+20
                 ----------------
                 d | (15n+21)-(15+20)=1 => d=1 => (5n+7) si (3n+4) sunt prime intre ele => fractia este ireductibila.