Hoặc
c) 1−y3y+1 . 5y+5y+2y+1;
c) 1−y3y+1 . 5y+5y+2y+1=−(y3−1)(5y+5)(y+1)(y+2y+1)
=−5(y−1)(y+2y+1)(y+1)(y+1)(y+2y+1)
=−5(y−1)=5−5y;