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Ma puteți ajuta la problema 6?

Ma Puteți Ajuta La Problema 6 class=

Răspuns :

E(x)=[tex] \frac{x}{ x^{2} (x+1)} : \frac{(x+2)(2x-1)-x(x+3)+1}{6(x+1)(x-1)} = \frac{1}{x(x+1)} * \frac{6(x+1)(x-1)}{(x+2)(2x-1)-x(x+3)+1} = \frac{6(x-1)}{x[2 x^{2} -x+4x-2- x^{2} -3x+1]} = \frac{6(x-1)}{x( x^{2} -1)} = \frac{6(x-1)}{x(x+1)(x-1)} = \frac{6}{x(x+1)} [/tex]