Question

subpunctiul b VA ROG​

Answer 1

Explicație pas cu pas:

x²+x = x(x+1)

x²+3x+2 = x²+x+2x+2 = x(x+1)+2(x+1) = (x+1)(x+2)

x²+5x+6 = x²+2x+3x+6 = x(x+2)+3(x+2) = (x+2)(x+3)

$E(x) = \dfrac{1}{x^{2}+x} + \dfrac{1}{x^{2}+3x+2} + \dfrac{1}{x^{2}+5x+6} =$

$= \dfrac{^{(x+2)(x+3)}1}{x(x+1)} + \dfrac{^{x(x+3)}1}{(x+1)(x+2)} + \dfrac{^{x(x+1)}1}{(x+2)(x+3)}$

$= \dfrac{(x+2)(x+3) + x(x+3) + x(x+1)}{x(x+1)(x+2)(x+3)}$

$= \dfrac{ {x}^{2} + 5x + 6 + {x}^{2} + 3x + {x}^{2} + x}{x(x+1)(x+2)(x+3)}$

$= \dfrac{ 3{x}^{2} + 9x + 6}{x(x+1)(x+2)(x+3)} = \dfrac{ 3({x}^{2} + 3x + 2)}{x(x+1)(x+2)(x+3)}$

$= \dfrac{ 3(x + 1)(x + 2)}{x(x+1)(x+2)(x+3)} = \dfrac{ 3}{x(x+3)}$

q.e.d.