a) x∈R\{-[tex] \frac{1}{2} [/tex];[tex] \frac{1}{2} [/tex];0;[tex] \frac{1}{8} [/tex]} cred ca asa este aici
b)E(x)=[tex]( \frac{2x-1+2x-1-6x}{4 x^{2}-1 }): \frac{4 x^{2}-1 }{8x}*x* \frac{4x}{x(x+3)}
[/tex]
E(x)=[tex] \frac{-2x}{4 x^{2}-1 }* \frac{4 x^{2}-1 }{8} * \frac{4}{x(x+3)} [/tex]
E(x)=[tex] \frac{-2x}{2x(x+3)} => E(x)= \frac{-1}{x+3} [/tex]
c) E([tex] \sqrt{3} [/tex]-3) = [tex] \frac{-1}{ \sqrt{3}-3+3 } = \frac{-1}{ \sqrt{3} } = \frac{- \sqrt{3} }{3} [/tex]
