Likninger med brøk

$\begin{array}{rcl}\frac{x}{2}+3 & =& \hspace{0.17em}1 |·2\\ x+6 & =& 2 |-6\\ x & =& -4\end{array}$

$\begin{array}{rcl}\frac{2x}{3}-\frac{1}{2} & =& x+2 |·2·3\\ 4x -3 & =& 6x +12 |-4x-12\\ -15 & =& 2x |:2\\ x & =& \hspace{0.17em}\frac{-15}{2}=-\frac{15}{2}\end{array}$

$\begin{array}{rcl}\frac{2}{3}+3x & =& \frac{1}{2}\left(x+2\right) |·2·3\\ 2·2 + 3x·2·3 & =& 3\left(x+2\right)\\ 4+18x & =& 3x+6 |-3x-4\\ 15x & =& 2 |:15\\ x & =& \frac{2}{15}\end{array}$

$\begin{array}{rcl}\frac{1}{2}x-\frac{1}{3}\left(x+2\right) & =& \frac{x}{4} |·2·2·3\\ 2·3·x-2·2\left(x+2\right) & =& x·3\\ 6x-4x-8 & =& \hspace{0.17em}3x |-2x\\ -8 & =& x\end{array}$

$\begin{array}{rcl}\frac{x}{4}+3 & =& 5x-\frac{1}{6} |·2·2·3\\ 3x+3·12 & =& \hspace{0.17em}5x·12-2\\ 3x+36 & =& 60x-2 |-3x+2\\ 38 & =& 57x |:57\\ x & =& \frac{38}{57}=\frac{38:19}{57:19}=\frac{2}{3}\\ & & \\ & & \end{array}$

$\begin{array}{rcl}3-4x & =& 2+\frac{x}{2} |·2\\ 6-8x & =& 4+x |+8x-4\\ 2 & =& 9x |:9\\ x & =& \hspace{0.17em}\frac{2}{9} \end{array}$

$\begin{array}{rcl}\frac{2+x}{2}+\frac{x}{3} & =& \frac{x}{6}-1 |·2·3\\ 3\left(2+x\right)+2x & =& x-6\\ 6+3x+2x & =& x-6\\ 6+5x & =& x-6 |-x-6\\ 4x & =& -12 |:4\\ x & =& -3 \end{array}$

Nevnerne er x og 3x slik at fellesnevner blir 3x.

$\begin{array}{rcl}\frac{1}{x}+\frac{2}{3x} & =& 5 |·3x\\ 3+2 & =& 15x\\ x & =& \frac{5}{15}=\frac{1}{3}\end{array}$

$\begin{array}{rcl}\frac{2}{x}-\frac{3}{4x} & =& \frac{5}{8} |·4·2·x\\ 2·4·2-3·2 & =& 5·x\\ 16-6 & =& 5x\\ x & =& \frac{10}{5}=2\end{array}$

$\begin{array}{rcl}\frac{2}{x}-5 & =& \frac{3}{2x}-3 |·2x\\ 4-10x & =& 3-6x |-3+10x\\ 1 & =& 4x |:4\\ x & =& \hspace{0.17em}\frac{1}{4}\end{array}$

$\begin{array}{rcl}\frac{1}{x-2}-\frac{x-4}{\left(x+2\right)\left(x-2\right)} & =& \frac{2}{x+2} |·\left(x+2\right)\left(x-2\right)\\ \frac{1·\left(x+2\right)\cancel{\left(x-2\right)}}{\cancel{\left(x-2\right)}}-\frac{\left(x-4\right)·\cancel{\left(x+2\right)\left(x-2\right)}}{\cancel{\left(x+2\right)\left(x-2\right)}} & =& \frac{2·\cancel{\left(x+2\right)}\left(x-2\right)}{\cancel{\left(x+2\right)}}\\ x+2-\left(x-4\right) & =& 2\left(x-2\right)\\ x+2-x+4 & =& 2x-4 |+4\\ 10 & =& 2x\\ x & =& 5\end{array}$

$\begin{array}{rcl}\frac{1}{x}+\frac{2}{x-3} & =& -\frac{3}{x} |·x\left(x-3\right)\\ x-3+2x & =& - 3\left(x-3\right)\\ 3x-3\hspace{0.17em}& =& -3x+9 |+3x+3\\ 6x & =& 12 |:6\\ x & =& \hspace{0.17em}2\end{array}$

$\begin{array}{rcl}\frac{2}{x-2}+\frac{4}{x-3} & =& \frac{1}{\left(x-2\right)\left(x-3\right)} |·\left(x-2\right)\left(x-3\right)\\ \frac{2·\cancel{\left(x-2\right)}\left(x-3\right)}{\cancel{\left(x-2\right)}}+\frac{4·\left(x-2\right)\cancel{\left(x-3\right)}}{\cancel{\left(x-3\right)}} & =& \frac{1·\cancel{\left(x-2\right)\left(x-3\right)}}{\cancel{\left(x-2\right)\left(x-3\right)}}\\ 2x-6+4x-8 & =& 1 |+14\\ 6x & =& 15 |:6\\ x & =& \frac{15}{6}=\frac{5}{2}\end{array}$

$\begin{array}{rcl}\frac{1}{2}\left(x+3\right)-\frac{x}{x-2} & =& \frac{1}{3}x+\frac{x}{6} |·2·3·\left(x-2\right)\\ \frac{1\cancel{·2}·3·\left(x-2\right)}{\cancel{2}}\left(x+3\right)-\frac{x·2·3·\cancel{\left(x-2\right)}}{\cancel{\left(x-2\right)}} & =& \frac{1·2·\cancel{3}·\left(x-2\right)}{\cancel{3}}x+\frac{x\cancel{·2·3}·\left(x-2\right)}{\cancel{6}}\\ 3\left(x-2\right)\left(x+3\right)-6x & =& 2\left(x-2\right)·x+x\left(x-2\right)\\ 3(x·x+3·x-2·x-2·3)-6x & =& 2·x·x-2·2·x+x·x-2·x\\ 3x^2+9x-6x-18-6x & =& 2x^2-4x+x^2-2x |-3x^2+6x+18\\ 3x & =& 18 |:3\\ x & =& 6\end{array}$

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

Her legger vi merke til at vi får 0 under brøkstreken hvis vi setter inn $x=1$, altså har likningen ingen løsninger.

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

$\begin{array}{rcl}1+\frac{2x}{x-2} & =& \frac{4}{x-2} |·\left(x-2\right)\\ x-2+2x & =& 4 |+2\\ 3x & =& 6 |:2\\ x & =& \hspace{0.17em}2\end{array}$

Her legger vi merke til at vi får 0 under brøkstreken hvis vi setter inn $x=2$, altså har likningen ingen løsninger.