1//(a+1)+(2a-10)//(a^2-a-2)+a//(a-2)

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вопрос

To simplify the given expression, we first need to factor the denominators of the fractions:

(a+1) = (a+2)(a-1)
(2a-10) = 2(a-5)
(a^2-a-2) = (a-2)(a+1)

Now, the given expression becomes:

[(a+1)/(a+2)(a-1)] + [(2a-10)/(a-2)] + [a/(a-2)]

To add these fractions, we need to find a common denominator which is the product of all the denominators:

Common denominator = (a+2)(a-1)(a-2)

Now, rewrite each fraction with the common denominator:

[(a+1)(a-2)/common denominator] + [(2(a-5)(a+2)/common denominator] + [a(a+2)/common denominator]

Now, we combine the numerators:

[(a^2-a-2) + 2(a-5)(a+2) + a(a+2)] / common denominator
= [a^2 - a - 2 + 2(a^2 - 3a - 10) + a^2 + 2a] / common denominator
= [a^2 - a - 2 + 2a^2 - 6a - 20 + a^2 + 2a] / common denominator
= [4a^2 - 5a - 22] / (a+2)(a-1)(a-2)

Therefore, the simplified expression is (4a^2 - 5a - 22) / ((a+2)(a-1)(a-2)).