СВ
Строганов Василий
ОТВЕТ: -(y+3)/5
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[(6 /(у^2-9))+1 /(3-у)] * (у^2 +6у+ 9) /5 =
= [(6/(y-3)(y+3)) - 1/(y-3)]*(y+3)^2/5 =
= [(6 - (y+3))/(y-3)(y+3)]*(y+3)^2/5 =
= [(3 - y)/(y-3)(y+3)]*(y+3)^2/5 =
=[-1/(y+3)]*(y+3)^2/5 =
= -(y+3)/5