Forenkling av rasjonale uttrykk

$$=\frac{16\left(x^2-4\right)}{4\left(x+2\right)}=\frac{4·\cancel{4}\cancel{\left(x+2\right)}\left(x-2\right)}{\cancel{4}\cancel{\left(x+2\right)}}=4x-8$$

$$=\frac{2\left(a^2-25\right)}{18\left(a-5\right)}=\frac{2\left(a^2-5^2\right)}{2·9\left(a-5\right)}=\frac{\cancel{2}\cancel{\left(a-5\right)}\left(a+5\right)}{\cancel{2}·9\cancel{\left(a-5\right)}}=\frac{a+5}{9}$$

$$=\frac{2\left(x^2-2^2\right)}{x-2}=\frac{2\cancel{\left(x-2\right)}\left(x+2\right)}{\cancel{x-2}}=2x+4$$

$$=\frac{\cancel{\left(x-{\displaystyle \frac{1}{2}}\right)}\left(x+{\displaystyle \frac{1}{2}}\right)}{2\cancel{\left(x-{\displaystyle \frac{1}{2}}\right)}}=\frac{x+{\displaystyle \frac{1}{2}}}{2}=\frac{x}{2}+\frac{1}{4}$$

$$=\frac{\cancel{2}\cancel{\left(x-{\displaystyle \frac{1}{3}}\right)}\left(x+{\displaystyle \frac{1}{3}}\right)}{\cancel{2}·3\cancel{\left(x-{\displaystyle \frac{1}{3}}\right)}}=\frac{x+{\displaystyle \frac{1}{3}}}{3}=\frac{3x+1}{9}=\frac{x}{3}+\frac{1}{9}$$

$$=\frac{-\left(-1+x\right)}{x-1}=-1$$

$$=\frac{-\left(-1+x^2\right)}{x^2-1}=-1$$

$$=\frac{-\cancel{\left(-1+x\right)}}{\cancel{\left(x-1\right)}\left(x+1\right)}=-\frac{1}{x+1}$$

$$=\frac{\left(x-1\right)\left(x-1+2\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{\cancel{\left(x-1\right)\left(x+1\right)}}{2\cancel{\left(x-1\right)\left(x+1\right)}}=\frac{1}{2}$$

$$\begin{array}{rcl}& = & \frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\\ & & \\ & =& \frac{2x+2-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ & & \\ & =& \frac{2x+2-x+1}{\left(x-1\right)\left(x+1\right)}\\ & & \\ & =& \frac{x+3}{\left(x-1\right)\left(x+1\right)}\end{array}$$

$$\begin{array}{rcl}& = & \frac{4\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\frac{8}{4x^2-1}\\ & & \\ & =& \frac{8x+4-8}{\left(2x-1\right)\left(2x+1\right)}\\ & & \\ & =& \frac{8x-4}{\left(2x-1\right)\left(2x+1\right)}\\ & & \\ & =& \frac{4\cancel{\left(2x-1\right)}}{\cancel{\left(2x-1\right)}\left(2x+1\right)}\\ & & \\ & =& \frac{4}{2x+1}\end{array}$$

$$\begin{array}{rcl}& = & \frac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3x+10}{\left(x-2\right)\left(x+2\right)}\\ & & \\ & =& \frac{4x+8-3x-10}{\left(x-2\right)\left(x+2\right)}\\ & & \\ & =& \frac{\cancel{x-2}}{\cancel{\left(x-2\right)}\left(x+2\right)}\\ & & \\ & =& \frac{1}{x+2}\end{array}$$

$$\begin{array}{rcl}& = & \frac{4}{2\left(x-5\right)}-\frac{{\displaystyle \frac{3x·\cancel{2}}{\cancel{2}}+\frac{15·\cancel{2}}{\cancel{2}}}}{2\left(x-5\right)\left(x+5\right)}\\ & & \\ & =& \frac{4}{2\left(x-5\right)}-\frac{3\cancel{\left(x+5\right)}}{2\left(x-5\right)\cancel{\left(x+5\right)}}\\ & & \\ & =& \frac{4-3}{2\left(x-5\right)}\\ & & \\ & =& \frac{1}{2\left(x-5\right)}\end{array}$$

CAS/GeoGebra: 4/(x-2)-(3x+10)/(x^2-4)

$$=\frac{1}{x+2}$$

CAS/GeoGebra: 4/(2x-10)-((3x)/2+15/2)/(x^2-25)

$$=\frac{1}{2\left(x-5\right)}$$