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\require{AMSmath}
Re: Re: Berekenen van onbekende a en b
Sorry maar wat u doet snap ik, maar de waarde van a en b berekenen lukt me nog niet.
mboudd
Leerling mbo - dinsdag 17 september 2019
Antwoord
Met behulp van substitutie:
$ \begin{array}{l} \left\{ \begin{array}{l} - \frac{1}{2}a + b = 0 \\ a = 2\sqrt b \\ \end{array} \right. \\ \left\{ \begin{array}{l} - \frac{1}{2} \cdot \left( {2\sqrt b } \right) + b = 0 \\ a = 2\sqrt b \\ \end{array} \right. \\ \left\{ \begin{array}{l} - \sqrt b + b = 0 \\ a = 2\sqrt b \\ \end{array} \right. \\ \left\{ \begin{array}{l} b = \sqrt b \\ a = 2\sqrt b \\ \end{array} \right. \\ \left\{ \begin{array}{l} b^2 = b \\ a = 2\sqrt b \\ \end{array} \right. \\ \left\{ \begin{array}{l} b^2 - b = 0 \\ a = 2\sqrt b \\ \end{array} \right. \\ \left\{ \begin{array}{l} b\left( {b - 1} \right) = 0 \\ a = 2\sqrt b \\ \end{array} \right. \\ \left\{ \begin{array}{l} b = 0 \\ a = 2\sqrt b \\ \end{array} \right. \vee \left\{ \begin{array}{l} b = 1 \\ a = 2\sqrt b \\ \end{array} \right. \\ \left\{ \begin{array}{l} b = 0 \\ a = 0 \\ \end{array} \right.(v.n.) \vee \left\{ \begin{array}{l} b = 1 \\ a = 2 \\ \end{array} \right. \\ \left\{ \begin{array}{l} a = 2 \\ b = 1 \\ \end{array} \right. \\ \end{array} $
Hoe moelijk kan dat zijn?
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dinsdag 17 september 2019
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