Во втором варианте 1 и 10, а в третьем варианте 10 и 8

1
$\frac{9}{14}: \frac{3}{7}+0.2= \frac{9}{14}* \frac{7}{3}+0.2= \frac{3}{2}+0.2=1.5+0.2=1.7$

10
$\frac{4}{y-3}+ \frac{6}{(y-3)^2} : \frac{6}{9-y^2}- \frac{2y-8}{y-3}= \frac{4}{y-3}+ \frac{6}{(y-3)^2} *\frac{-(y+3)(y-3)}{6}- \frac{2y-8}{y-3}= \\\\ = \frac{4}{y-3}+ \frac{-y-3}{y-3} - \frac{2y-8}{y-3}=\frac{4-y-3-2y+8}{y-3}=\frac{9-3y}{y-3}=\frac{-3(y-3)}{y-3}=-3$

8
$y= \frac{1}{2} x+1$

10
$2y^3-3x^3+x^2y^2-6xy=2y^3-6xy+x^2y^2-3x^3= \\ =2y(y^2-3x)+x^2(y^2-3x)=(y^2-3x)(2y+x^2)$