Wiskundeleraar

Uitwerkingen

Opgave 1

$$
\eqalign{
& z^2 + z^2 = \left( {6\sqrt 3 } \right)^2 \cr
& 2z^2 = 108 \cr
& z^2 = 54 \cr
& z = \pm 3\sqrt 6 \cr}
$$

$$
\eqalign{
& z^2 + z^2 = \left( {10\sqrt 5 } \right)^2 \cr
& 2z^2 = 500 \cr
& z^2 = 250 \cr
& z = \pm 5\sqrt {10} \cr}
$$

$$
\eqalign{
& z^2 + z^2 = \left( {14\sqrt 7 } \right)^2 \cr
& 2z^2 = ... \cr
& z^2 = ... \cr
& z = \pm 7\sqrt {14} \cr}
$$

Opgave 2

$$
\frac{1}
{{\sqrt 3 }} = \frac{{AB}}
{{6\sqrt 2 }} \Rightarrow AB = 2\sqrt 6
$$

$$
\frac{1}
{{\sqrt 3 }} = \frac{{AB}}
{{15\sqrt 5 }} \Rightarrow AB = 5\sqrt {15}
$$

$$
\frac{1}
{{\sqrt 3 }} = \frac{{AB}}
{{39\sqrt {13} }} \Rightarrow AB = 13\sqrt {39}
$$

Opgave 3

$\sqrt{3}+\sqrt{2}=\sqrt{5}$

$\sqrt{3}\times\sqrt{2}=\sqrt{6}$

$2\times\sqrt{3}=\sqrt{12}$

$\sqrt{8}+\sqrt{18}=\sqrt{50}$

$\sqrt { - 7} = - \sqrt 7$

$\sqrt{7}\times-\sqrt{7}=7$

$ \frac{1}{2}\sqrt{2}=\sqrt{\frac{1}{2}}$

$\sqrt{12}+\sqrt{27}=\sqrt{75}$