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Bruk tredje kvadratsetning nár du forkorter brokene. \( \begin{array}{lll}\text { a) } \frac{a^{2}-36}{(a+6)(a-6)} & \text { b) } \frac{x+7}{x^{2}-7^{2}} & \text { c) } \frac{2 x^{2}-50}{2 x+10}\end{array} \)

Ask by Ruiz Schmidt. in Norway
Mar 08,2025

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**a) 1 (gjelder for \(a \neq \pm6\))** **b) \(\frac{1}{x-7}\) (gjelder for \(x \neq -7\))** **c) \(x-5\) (gjelder for \(x \neq -5\))**

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Når vi bruker tredje kvadratsetning, kan vi forenkle brøker som inneholder kvadratiske uttrykk. For eksempel, for \(a^2 - 36\), ser vi at dette kan skrives som \((a + 6)(a - 6)\). Dermed kan brøken \( \frac{a^{2}-36}{(a+6)(a-6)} \) forenkles til 1, så lenge \(a \neq 6\) og \(a \neq -6\). I brøken \( \frac{x+7}{x^{2}-7^{2}} \), kan vi gjenkjenne \(x^2 - 7^2\) som \((x - 7)(x + 7)\). Det gir oss \( \frac{x+7}{(x-7)(x+7)} \), som forenkles til \( \frac{1}{x - 7} \), så lenge \(x \neq 7\). For den siste brøken, \( \frac{2x^{2}-50}{2x+10} \), kan vi faktorisere telleren til \(2(x^{2}-25)\), som er \(2(x+5)(x-5)\). Dermed kan vi forenkle brøken til \( \frac{2(x+5)(x-5)}{2(x+5)} \), som gir \(x - 5\) så lenge \(x \neq -5\).

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