Forenkling av rasjonale uttrykk

$=\frac{\cancel{\left(x+5\right)}\left(x-5\right)}{\cancel{\left(x+5\right)}}=x-5$

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

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

$=\frac{\left(10x+1\right)\cancel{\left(10x-1\right)}}{\cancel{\left(10x-1\right)}}=10x+1$

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

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

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

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

$=\frac{3\left(x^2-6x+9\right)}{2\left(x-3\right)}=\frac{3\cancel{\left(x-3\right)}\left(x-3\right)}{2\cancel{\left(x-3\right)}}=\frac{3\left(x-3\right)}{2}$

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

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

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

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

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

Her kan vi setje $\left(x-1\right)$ utanfor parentes sidan uttrykket er felles faktor i begge ledda i teljaren.

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

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

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

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

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

Her kan vi bruke knappen for faktorisering etterpå (sjå line 2 i CAS-biletet nedanfor) for å få det forenkla uttrykket på same form som i oppgåve 1.5.11 b).

Opne bilete i eit nytt vindauge

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Bilete: Bjarne Skurdal