помогите, пожалуйста, с номером 4.33. Created by daschatatarova. algebra-ru

помогите, пожалуйста, с номером 4.33...

daschatatarova @daschatatarova

August 2021 1 8 Report

помогите, пожалуйста, с номером 4.33

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Regent1828

Verified answer

a/(a²-1) + (a²+a-1) / (a³-a²+a-1) + (a²-a-1) / (a³+a²+a+1) - 2a³/ ((a²-1)(a²+1)) =
= a/(a-1)(a+1) + (a²+a-1)/(a²(a-1)+(a-1)) + (a²-a-1)/(a²(a+1)+(a+1)) - 2a³/((a²-1)(a²+1)) = a/(a-1)(a+1) + (a²+a-1)/(a²+1)(a-1) + (a²-a-1)/(a²+1)(a+1) - 2a³/((a²-1)(a²+1)) =
= a(a²+1)(a²-1)+(a²+a-1)(a²-1)(a+1)+(a²-a-1)(a²-1)(a-1)-2a³(a²-1) /
/ (a²+1)(a²-1)(a-1)(a+1) =
= a(a²+1)+(a²+a-1)(a+1)+(a²-a-1)(a-1)-2a³ / (a²+1)(a²-1) =
= a³+a+a³+a²+a²+a-a-1+a³-a²-a²+a-a+1-2a³ / (a²+1)(a²-1) =
= (a³+a) / (a²+1)(a²-1) =
= a(a²+1) / (a²+1)(a²-1) = a/(a²-1)

при a=3/2 3/2 : (9/4 -1) = 3/2 × 4/5 = 6/5 = 1,2

1 votes Thanks 2

Regent1828 Привет из Мурманска...)))

daschatatarova спасибо большое за решение)))

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