Question

Решите плиз

 

(6/(y^2-9)+1/(3-y))*((y^2+6*y+9)/5)

 

Answer 1

$(frac {6}{y^2-9}+frac {1}{3-y}) cdot (frac {y^2+6y+9}{5}) \ \ \ \ 1) frac {6}{y^2-9}+frac {1}{3-y}=frac {6}{(y-3)(y+3)}+frac {1}{3-y}=frac {6}{(y-3)(y+3)}-frac {1}{y-3}= \ \ =frac {6}{(y-3)(y+3)}-frac {3+y}{(y-3)(y+3)}=frac {6-(3+y)}{(y-3)(y+3)}=frac {6-3-y}{(y-3)(y+3)}=\ \ =frac {3-y}{(y-3)(y+3)}= -frac {1}{3+y} \ \ \ 2) -frac {1}{3+y} cdot frac {y^2+6y+9}{5}=- frac {1}{3+y} cdot frac {(y+3)^2}{5}=-frac {1 cdot (y+3)(y+3)}{(3+y) cdot 5}=-frac {y+3}{5}$

 

 

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