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a fost răspuns

E(x)=(x+1+[tex] \frac{2}{x-1} [/tex]) · [tex] \frac{2}{x-1} \frac{ x^{2}-2x+1}{ x ^{3}- x^{2} +x -1 } [/tex]

Răspuns :

Renatemambouko
2/(x-1) × (x²-2x+1)/(x³-x²+x-1)=
=2/(x-1) × (x-1)²/[x²(x-1)+(x-1)]=
=2 × (x-1)/(x-1)(x²+1)=
=2 /(x²+1)

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