Question

срочно решите прошу все

Answer 1

$1)\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}+\frac{a^{3}+b^{3}}{a^{2}-ab+b^{2}} =\frac{(a-b)(a^{2}+ab+b^{2})}{a^{2}+ab+b^{2}}+\frac{(a+b)(a^{2}-ab+b^{2})}{a^{2}-ab+b^{2}}=\\\\=a-b+a+b=2a\\\\a=0,05 \ \Rightarrow \ 2a=2*0,05=0,1\\\\Otvet:\boxed{0,1}$

$2)\frac{2}{x-2}-\frac{12x}{x^{3}-8 } -\frac{x-2}{x^{2}+2x+4 }=\frac{2}{x-2}-\frac{12x}{(x-2)(x^{2}+2x+4)} -\frac{x-2}{x^{2}+2x+4 } =\\\\=\frac{2*(x^{2}+2x+4)-12x-(x-2)(x-2)}{(x-2)(x^{2}+2x+4)}=\frac{2x^{2}+4x+8-12x-x^{2}+4x-4}{(x-2)(x^{2}+2x+4)}=\\\\=\frac{x^{2}-4x+4 }{(x-2)(x^{2}+2x+4)}=\frac{(x-2)^{2} }{(x-2)(x^{2}+2x+4)}=\boxed{\frac{x-2}{x^{2}+2x+4 }}\\\\1)x=0\\\\\frac{0-2}{0^{2}+2*0+4 }=\frac{-2}{4}=\boxed{-0,5}\\\\2)x=2\\\\\frac{2-2}{2^{2} +2*2+4}=\frac{0}{12}=\boxed0$