Dạng 1: Thực hiện phép tính

$\frac{1}{x^2+3x+2}-\frac{2x}{x^3+4x^2+4x}+\frac{1}{x^2+5x+6}$

$\begin{array}{l}=\frac{1}{x^2+2x+x+2}-\frac{2x}{x(x^2+4x+4)}+\frac{1}{x^2+3x+2x+6}\\ =\frac{1}{x(x+2)+(x+2)}-\frac{2x}{x{(x+2)}^2}+\frac{1}{x(x+3)+2(x+3)}\\ =\frac{1}{(x+2)(x+1)}-\frac{2x}{x{(x+2)}^2}+\frac{1}{(x+3)(x+2)}\\ =\frac{x(x+2)(x+3)}{x(x+1){(x+2)}^2(x+3)}-\frac{2x(x+1)(x+3)}{x(x+1){(x+2)}^2(x+3)}+\frac{x(x+1)(x+2)}{x(x+1){(x+2)}^2(x+3)}\\ =\frac{x(x+2)(x+3)-2x(x+1)(x+3)+x(x+1)(x+2)}{x(x+1){(x+2)}^2(x+3)}\\ =\frac{x^3+5x^2+6x-2x^3-8x^2-6x+x^3+3x^2+2x}{x(x+1){(x+2)}^2(x+3)}\\ =\frac{2x}{x(x+1){(x+2)}^2(x+3)}\\ =\frac{2}{(x+1){(x+2)}^2(x+3)}.\end{array}$

$\begin{array}{l}=\frac{1}{2x+3}-\frac{1}{2x-3}+\frac{x-2}{2x^2-x-3}\\ =\frac{1}{2x+3}-\frac{1}{2x-3}+\frac{x-2}{2x^2+2x-3x-3}\\ =\frac{1}{2x+3}-\frac{1}{2x-3}+\frac{x-2}{2x(x+1)-3(x+1)}\\ =\frac{(2x-3)(x+1)}{(2x-3)(2x+3)(x+1)}-\frac{(2x+3)(x+1)}{(2x-3)(2x+3)(x+1)}+\frac{(x-2)(2x+3)}{(2x-3)(2x+3)(x+1)}\\ =\frac{(2x-3)(x+1)-(2x+3)(x+1)+(x-2)(2x+3)}{(2x-3)(2x+3)(x+1)}\\ =\frac{2x^2-x-3-2x^2-5x-3+2x^2-x-6}{(2x-3)(2x+3)(x+1)}=\frac{2x^2-7x-12}{(2x-3)(2x+3)(x+1)}.\end{array}$

$\begin{array}{l}=\frac{1}{x^2+2x-x-2}+\frac{1}{x^2-2x+x-2}+\frac{1+x}{x^2+2x+1-x-3}\\ =\frac{1}{x(x+2)-(x+2)}+\frac{1}{x(x-2)+(x-2)}+\frac{1+x}{x^2+2x-x-2}\\ =\frac{1}{(x-1)(x+2)}+\frac{1}{(x+1)(x-2)}+\frac{1+x}{(x-1)(x+2)}\\ =\frac{(x+1)(x-2)}{(x-1)(x+1)(x-2)(x+2)}+\frac{(x-1)(x+2)}{(x-1)(x+1)(x-2)(x+2)}+\frac{(1+x)(x+1)(x-2)}{(x-1)(x+1)(x-2)(x+2)}\\ =\frac{x^2-x-2+x^2+x-2+x^3-3x-2}{(x-1)(x+1)(x-2)(x+2)}\\ =\frac{x^3+2x^2-3x-6}{(x-1)(x+1)(x-2)(x+2)}\\ =\frac{(x^2-3)(x+2)}{(x-1)(x+1)(x-2)(x+2)}\\ =\frac{x^2-3}{(x-1)(x+1)(x-2)}.\end{array}$

$\begin{array}{l}=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\\ =\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\\ =\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\\ =\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\\ =\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\\ =\frac{32}{1-x^{32}}.\end{array}$