Question

Выражения и найти их значения:

Answer 1

$\bigg(\big(\frac{n+2}{n-2})^3 \cdot\frac{n^3+4n^2+4n}{3n^2-12n+12}\bigg)\cdot\frac{n}{3}=\frac{(n+2)^3}{(n-2)^3} \cdot\frac{n(n^2+4n+4)}{3(n^2-4n+4)}\cdot\frac{n}{3}=\\\\=\frac{(n+2)^3\cdot n(n+2)^2\cdot n}{(n-2)^3\cdot 3(n-2)^2\cdot 3}=\frac{n^2(n+2)^5}{9(n-2)^5}=\frac{n^2}{9}(\frac{n+2}{n-2})^5 \\\\ \frac{n^2}{9}(\frac{n+2}{n-2})^5=\frac{(-0,5)^2}{9}(\frac{-0,5+2}{-0,5-2})^5=\frac{0,25}{9}(\frac{1,5}{-2,5})^5=\\\\ \frac{1}{9\cdot4}(-\frac{3}{5})^5=-\frac{3^5}{3^2\cdot4\cdot5^5}=\frac{27}{12500}=-0,00216$

$4x^3-8x^2+2x+3=4x^2(x-2)+2x+3\\ \\ 4(\frac{1}{2}(1+\sqrt{3} ))^2(\frac{1}{2}(1+\sqrt{3} )-2)+2\cdot\frac{1}{2}(1+\sqrt{3} )+3=\\ \\ =(1+\sqrt{3})^2(\frac{1}{2}+\frac{\sqrt{3}}{2}-2)+(1+\sqrt{3} )+3=(1+2\sqrt{3}+3)(\frac{\sqrt{3}}{2}-\frac{3}{2} )+4+\sqrt{3}=\\ \\ =(2+\sqrt{3})(\sqrt{3}-3)+4+\sqrt{3}=2\sqrt{3}+3-6-3\sqrt{3}+4+\sqrt{3}=1$