Помогите пожалуйста с заданиями : ж), 8(а, б, в), 9

Ж) $( \frac{3-a}{a^2-2a+1} - \frac{2}{1-a} )*( \frac{a^2-3a}{a^3+3a^2+3a+1} + \frac{1}{a^2+2a+1} )=$
$= (\frac{3-a}{(a-1)^2}+ \frac{2}{a-1} )*( \frac{a^2-3a}{(a+1)^3} + \frac{1}{(a+1)^2} )= \frac{3-a+2(a-1)}{(a-1)^2}* \frac{a^2-3a+a+1}{(a+1)^3}=$
$= \frac{3-a+2a-2}{(a-1)^2}* \frac{a^2-2a+1}{(a+1)^3} = \frac{a+1}{(a-1)^2}* \frac{(a-1)^2}{(a+1)^3}= \frac{1}{(a+1)^2}$
ОДЗ: a =/= -1; a =/= 1

8) x^2 - 10x + 12 = 0
x1 + x2 = -b/a = 10; x1*x2 = c/a = 12

a) $\frac{x2}{x1}+ \frac{x1}{x2}= \frac{x2^2+x1^2}{x1*x2}= \frac{x1^2+2x1*x2+x2^2-2x1*x2}{x1*x2} =$
$=\frac{(x1+x2)^2}{x1*x2}-2= \frac{10^2}{12}-2= \frac{100}{12}-2= \frac{25}{3}-2= \frac{19}{3}$

б) 3x1+3x2-4x1*x2=3(x1+x2)-4*x1*x2 = 3*10 - 4*12 = -18

в) $x2 = \frac{-b+ \sqrt{D} }{2a}; x1= \frac{-b- \sqrt{D} }{2a}$
$D=10^2-4*12=100-48=52[tex]|x2-x1|=|\frac{-b+ \sqrt{D} }{2a}-\frac{-b- \sqrt{D} }{2a}|=| \frac{ \sqrt{D} }{2a}+ \frac{ \sqrt{D} }{2a} |=| \frac{ \sqrt{D} }{a} |= \frac{ \sqrt{52} }{1}=2 \sqrt{13}$ [/tex]

9)
{ x^2 + y^2 = 6
{ xy = 2

x^4 + x^2*y^2 + y^4 = x^4 + 2x^2*y^2 + y^4 - x^2*y^2 =
= (x^2 + y^2)^2 - (xy)^2 =6^2 - 2^2 = 36 - 4 = 32