1.4 Lösning 12

\(\begin{align} v - {u \over u\,v + v\,x} & = {v\,x^2 \over x^2 - u^2} + {u\,v^2 \over v\,x + u\,v} \\ v - {u \over v\,(u + x)} & = {v\,x^2 \over (x+u)\,(x-u)} + {u\,v^2 \over v\,(x + u)} \\ v - {u \over v\,(x + u)} & = {v\,x^2 \over (x+u)\,(x-u)} + {u\,v^2 \over v\,(x + u)} & | \;\cdot v\,(x + u)\,(x-u)\\ v^2\,(x + u)\,(x-u) - u\,(x-u) & = v^2\,x^2 + u\,v^2\,(x-u) \\ v^2\,(x^2-u^2) - u\,(x-u) & = v^2\,x^2 + u\,v^2\,(x-u) \\ v^2\,x^2-v^2\,u^2 - u\,x + u^2 & = v^2\,x^2 + u\,v^2\,x - u\,v^2 \\ \end{align}\)

\(\begin{align} \sqrt{6 x + 10} + 1 & = x & | \;\; -1 \\ \sqrt{6 x + 10} & = x - 1 & | \; (\;\;\;)^2 \\ 6 x + 10 & = (x - 1)^2 \\ 6 x + 10 & = x^2 - 2 x + 1 \qquad\qquad & | - 10 \\ 6 x & = x^2 - 2 x - 9 \qquad\qquad & | - 6 x \\ 0 & = x^2 - 8 x - 9 \\ \end{align}\)

Exempel 6

Förenkla följande uttryck så långt som möjligt\[ {2 \over x^2-4} \, + \, {1 \over 2\,x - x^2} \]

\( {2 \over x^2-4} \, + \, {1 \over 2\,x - x^2} \; = \; {2 \over (x+2)\cdot(x-2)} \, + \, {1 \over (2-x)\cdot x} \; = \; {2 \over (x+2)\cdot(x-2)} \, + \, {1 \, \over - \, (x-2)\cdot x} \; = \; \)

\( = \; {2 \over (x+2)\cdot(x-2)} \, + \, {-1 \over (x-2)\cdot x} \; = \; {\qquad\quad 2 \qquad\quad\;\cdot {\color{Red} x} \over (x+2)\cdot(x-2) \cdot {\color{Red} x}} \; + \; {{\color{Red} (x+2)}\cdot \quad\, (-1) \quad\, \over {\color{Red} (x+2)}\cdot (x-2)\cdot x} \; = \; \)

\( = \; {2\,x \; + \; (x+2) \cdot (-1) \over (x+2) \cdot (x-2)\cdot x} \; = \; {2\,x \; + \; (-x-2) \over (x+2) \cdot (x-2)\cdot x} \; = \; {2\,x - x - 2 \over (x+2) \cdot (x-2)\cdot x} \; = \)

\( = \; {x - 2 \over (x+2) \cdot (x-2)\cdot x} \; = \; {1 \over x \; (x+2)} \)