[Risolto] Frazioni algebriche

1/(a^2 - 2·a + 1) + 7/(2·a - 2) + 1/(2·a + 2) + 2·a + 2=

=1/(a - 1)^2 + 7/(2·(a - 1)) + 1/(2·(a + 1)) + 2·a + 2=

C.E. 2·(a - 1)^2·(a + 1) ≠ 0---> a ≠ -1 ∧ a ≠ 1

=(1·2·(a + 1) + 7·(a - 1)·(a + 1) + 1·(a - 1)^2)/(2·(a - 1)^2·(a + 1)) + 2·a + 2=

=((2·a + 2) + (7·a^2 - 7) + (a^2 - 2·a + 1))/(2·(a - 1)^2·(a + 1)) + 2·a + 2=

=(8·a^2 - 4)/(2·(a - 1)^2·(a + 1)) + 2·a + 2=

=2·a^4/((a + 1)·(a - 1)^2)

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1/(x^2 - 1) - 2/(x^2 + 3·x - 4) + 1/(x^2 - 2·x - 3)=

=1/((x + 1)·(x - 1)) - 2/((x - 1)·(x + 4)) + 1/((x + 1)·(x - 3))=

C.E. (x + 1)·(x - 1)·(x + 4)·(x - 3) ≠ 0----> x ≠ -4 ∧ x ≠ 3 ∧ x ≠ -1 ∧ x ≠ 1

=(1·(x + 4)·(x - 3) - 2·(x + 1)·(x - 3) + 1·(x - 1)·(x + 4))/((x + 1)·(x - 1)·(x + 4)·(x - 3)) =

=((x^2 + x - 12) - (2·x^2 - 4·x - 6) + (x^2 + 3·x - 4))/((x + 1)·(x - 1)·(x + 4)·(x - 3))=

=(8·x - 10)/((x + 1)·(x - 1)·(x + 4)·(x - 3))=

=(8·x - 10)/(x^4 + x^3 - 13·x^2 - x + 12)