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

ж) [latex]( \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} )=[/latex] [latex]= (\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}= [/latex] [latex]= \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} [/latex] ОДЗ: a =/= -1; a =/= 1 8) x^2 - 10x + 12 = 0 x1 + x2 = -b/a = 10; x1*x2 = c/a = 12 a) [latex] \frac{x2}{x1}+ \frac{x1}{x2}= \frac{x2^2+x1^2}{x1*x2}= \frac{x1^2+2x1*x2+x2^2-2x1*x2}{x1*x2} = [/latex] [latex]=\frac{(x1+x2)^2}{x1*x2}-2= \frac{10^2}{12}-2= \frac{100}{12}-2= \frac{25}{3}-2= \frac{19}{3} [/latex] б) 3x1+3x2-4x1*x2=3(x1+x2)-4*x1*x2 = 3*10 - 4*12 = -18 в) [latex]x2 = \frac{-b+ \sqrt{D} }{2a}; x1= \frac{-b- \sqrt{D} }{2a}[/latex] [latex]D=10^2-4*12=100-48=52[latex]|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} [/latex][/latex] 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