помогите решить срочно.... Created by lapochka1. algebra-ru

помогите решить срочно...

karasikov $(a^{2} +12a+\frac{64}{a} +48)* \frac{1}{a^2-16}*(a^2-4a)=$ $\frac{a^3+12a^2+48a+64}{a}* \frac{1}{(a-4)(a+4)}*a(a-4) = \frac{a^3+12a^2+48a+64}{a+4}$ = $\frac{(a+4)^3}{a+4}=(a+4)^2$

1 votes Thanks 2

alika7 =(a^3+12a^2+64+48a)*(1/(a-4)(a+4))*(a(a-4))= ((a^3+64)+(12a^2+48a))*(a/(a+4))=
=((a+4)(a^2-4a+16)+12a(a+4)) *(a/(a+4))= (a+4)(a^2-4a+12a+16)*a/(a+4)=
=(a^2+8a+16)*a=a^3+8a^2+16a

2 votes Thanks 1

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