Question

Ajutati-ma.... la expresia asta de groaza

Answer 1

a)  [ (x+1) / x(x-2) + x / (x-2)(x+2) - (x+1) / x(x+2)] ·x(x-2)² / (x+3)(x+2) =
= [(x+1)(x+2) + x² - (x+1)(x-2)] /x(x+2)(x-2)· x(x-2)² / (x+3)(x+2) =
= ( x² + 3x + 2 + x² - x²+x +2) / x(x+2)(x-2) · x(x-2)² / (x+3)(x+2) =
= (x² +4x+4) / (x+2) · (x-2) /(x+3)(x+2) = (x+2)² /(x+2) ·(x-2) / (x+3)(x+2) = (x-2) /(x+3)
E(x) = (x-2) / (x+3)
pentru ca E(x)∈ Z trebuie ca  (x+3) sa divida (x-2)
presupunem ca  d divide (x+ 3)      (1)
d divide (x-2)      (2) ⇒      d divide (1) - (2) = 5
divizorii lui 5 ∈ {-5, -1, 1, 5}
1) daca x+3 = - 5    x = -8    E(-8) = -10/-5 = 2 ∈ Z 
2) daca x+3 = - 1    x = -4    E(-4) = -6/ -1 = 6 ∈ Z
3)  daca x+3 = 1      x = - 2    E(-2) = -4/1 = -4 ∈ Z
4) daca x+3 = 5    x = 2      E(2) = 0 ∈Z
b)  F(x) = (x-2)/(x+3) · (x+3) = x-2
F(3) + F(4) +.........+ F(2015) = 1 + 2 + 3 + .........2013 = 2013·2014/2 =
= 2013·1007 = 2027091