Помогите решить плиз.... даю 25б

1) $(a-4)^{2}-2a(3a-4)=a^{2}-8a+16-6a^{2}+8a=-5a^{2}+16$
$\frac{c^{7}*c^{3}}{(c^{2})^{4}} = \frac{c^{10}}{c^{8}}= c^{2}$
$\frac{x^{2}+xy}{x^{2}-y^{2}}= \frac{x(x+y)}{(x+y)(x-y)}= \frac{x}{x-y}$
2)пересечения (1;-4)
3) $\left \{ {{x+8y=-6} \atop {5x-2y=12}} \right. \\ x=-6-8y \\ 5*(-6-8y)-2y=12 \\ -30-40y-2y=12 \\ -42y=12+30 \\ -42y=42 \\y=-1 \\x=-6-8y \\x=-6-8*(-1) \\x=-6+8 \\x=2$
1) $(x-y)^{2}-x(x-2y)=x^{2}-2xy+y^{2}-x^{2}+2xy=y^2$
$\frac{a^{5}*a^{8}}{(a^{3})^{4}}=\frac{a^{13}}{a^{12}}=a$
$\frac{m^{2}-4}{m^2-2m}=\frac{(m+2)(m-2)}{m(m-2)}=\frac{m+2}{m}$
2)пересечение (1;4)
3) $\left \{ {{4x-y=9} \atop {3x+7y=-1}} \right. \\-y=9-4x =\ \textgreater \ y=4x-9 \\3x+7y=-1\\3x+7(4x-9)=-1\\3x+28x-63=-1\\31x=-1+63\\31x=62\\x=2\\y=4x-9\\y=4*2-9\\y=8-9\\y=-1$