Сроочно!! Ребята помогите пожалуйста! Вообще не могу решить(

[latex]d)\;\left(x-\frac{x^3+8}{2x+x^2}\right)\cdot\frac x{(x-2)^2}+\frac2{2-x}\\1)\;x-\frac{x^3+8}{2x+x^2}=x-\frac{(x+2)(x^2-2x+4)}{x(2+x)}=x-\frac{x^2-2x+4}{x}=\\=\frac{x^2-x^2+2x-4}{x}=\frac{2(x-2)}x\\2)\;\frac{2(x-2)}x\cdot\frac{x}{(x-2)^2}=\frac2{x-2}\\3)\;\frac2{x-2}+\frac2{2-x}=\frac2{x-2}-\frac2{x-2}=0[/latex] [latex]10.\;\frac{a^2+a-2}{a^4-3a^3}\cdot\left(\frac{(a+2)^2-a^2}{4a^2-4}-\frac3{a^2-a}\right)-\frac2{a^4}\\1)\;\frac{(a+2)^2-a^2}{4a^2-4}-\frac3{a^2-a}=\frac{a^2+4a+4-a^2}{4(a^2-1)}-\frac3{a(a-1)}=\\=\frac{4(a+1)}{4(a-1)(a+1)}-\frac3{a(a-1)}=\frac1{a-1}-\frac3{a(a-1)}=\frac{a-3}{a(a-1)}\\2)\;\frac{a^2+a-2}{a^4-3a^3}\cdot\frac{a-3}{a(a-1)}=\frac{a^2+a-2}{a^3(a-3)}\cdot\frac{a-3}{a(a-1)}=\frac{a^2+a-2}{a^4(a-1)}[/latex] [latex]3)\;\frac{a^2+a-2}{a^4(a-1)}-\frac2{a^4}=\frac{a^2+a-2-2(a-1)}{a^4(a-1)}=\frac{a^2+a-2-2a+2}{a^4(a-1)}=\frac{a^2-a}{a^4(a-1)}=\\=\frac{a(a-1)}{a^4(a-1)}=\frac1{a^3}\\a=\sqrt[3]2\\\frac1{a^3}=\frac1{(\sqrt[3]2)^3}=\frac12[/latex]

[latex](x- \frac{ x^{3}+8 }{2x+ x^{2} } )* \frac{x}{ (x-2)^{2} } +\frac{2}{x-2}= \frac{x(2x+x^{2})- x^{3}-8 }{x(2+x)}* \frac{x}{ (x-2)^{2} }- \frac{2}{x-2}= \\ = \frac{x(2 x^{2} + x^{3}- x^{3}-8 }{x( x^{2} -4)(x-2)} - \frac{2}{2-x} = \frac{2 x^{3}-8x-2x( x^{2} -4) }{x( x^{2} -4)(x-2)} = \frac{x(2 x^{2} -8-2 x^{2} +8)}{x( x^{2} -4)(x-2)} = \\ = \frac{x(0)}{x( x^{2} -4)(x-2)} =0[/latex]