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$100 \cdot \sqrt{\frac{30 miljoen}{10 miljoen}} = 100 \cdot \sqrt{3} \approx 173,2$ , dus $173$ afgevaardigden ;
$100 \cdot \sqrt{\frac{2 miljoen}{10 miljoen}} = 100 \cdot \sqrt{0,2} \approx 44,7$ , dus $45$ afgevaardigden

$100 \cdot \sqrt{\frac{n}{10}} = 100 \cdot \sqrt{\frac{10 n}{100}} = 100 \cdot \frac{1}{10} \sqrt{10 n} = 10 \sqrt{10 n}$ afgevaardigden

Lengte hoogtelijn is $\sqrt{{(2 \sqrt{6})}^{2} - {(2 \sqrt{2})}^{2}} = \sqrt{24 - 8} = 4$ ,
oppervlakte driehoek is $4 \cdot 2 \sqrt{2} = 8 \sqrt{2}$ .

$\sqrt{2 \frac{1}{2}} = \sqrt{\frac{10}{4}} = \sqrt{\frac{1}{4}} \cdot \sqrt{10} = \frac{1}{2} \sqrt{10}$

${(3 x)}^{2} = 12$		eerste manier $9 x^{2} = 12 \Leftrightarrow x^{2} = \frac{4}{3} \Leftrightarrow x = \sqrt{\frac{4}{3}}$ of $x = ‐ \sqrt{\frac{4}{3}}$ , $\sqrt{\frac{4}{3}} = \sqrt{4} \cdot \sqrt{\frac{1}{3}} = 2 \cdot \sqrt{\frac{3}{9}} = \frac{2}{3} \sqrt{3}$ , dus $x = \frac{2}{3} \sqrt{3}$ of $x = ‐ \frac{2}{3} \sqrt{3}$ .
${(3 x)}^{2} = 12$		tweede manier $3 x = \sqrt{12}$ of $3 x = ‐ \sqrt{12}$ , dus $x = \frac{1}{3} \sqrt{12} = \frac{2}{3} \sqrt{3}$ of $x = ‐ \frac{1}{3} \sqrt{12} = ‐ \frac{2}{3} \sqrt{3}$

${(3 x)}^{2} = 11$

eerste manier
$9 x^{2} = 11$
$x^{2} = \frac{11}{9}$
$x = \sqrt{\frac{11}{9}} = \frac{1}{3} \sqrt{11}$ of $x = ‐ \sqrt{\frac{11}{9}} = ‐ \frac{1}{3} \sqrt{11}$

${(3 x)}^{2} = 11$

tweede manier
$3 x = \sqrt{11}$ of $3 x = ‐ \sqrt{11}$ , dus
$x = \frac{1}{3} \sqrt{11}$ of $x = ‐ \frac{1}{3} \sqrt{11}$

${(\sqrt{3} x)}^{2} = 12$

eerste manier
$3 x^{2} = 12$
$x^{2} = 4$
$x = 2$ of $x = ‐ 2$

${(\sqrt{3} x)}^{2} = 12$

tweede manier
$\sqrt{3} x = \sqrt{12}$ of $\sqrt{3} x = ‐ \sqrt{12}$
$x = \frac{\sqrt{12}}{\sqrt{3}} = \sqrt{4} = 2$ of $x = ‐ 2$

${(\sqrt{3} x)}^{2} = 16$

eerste manier
$3 x^{2} = 16$
$x^{2} = \frac{16}{3}$
$x = \sqrt{\frac{16}{3}} = 4 \sqrt{\frac{1}{3}} = \frac{4}{3} \sqrt{3}$ of $x = ‐ \frac{4}{3} \sqrt{3}$

${(\sqrt{3} x)}^{2} = 16$

tweede manier
$\sqrt{3} x = 4$ of $\sqrt{3} x = ‐ 4$
$x = \frac{4}{\sqrt{3}} = \frac{4 \sqrt{3}}{\sqrt{9}} = \frac{4}{3} \sqrt{3}$ of $x = ‐ \frac{4}{3} \sqrt{3}$

${(\frac{x}{\sqrt{2}})}^{2} = \frac{1}{4}$

eerste manier
$\frac{x^{2}}{2} = \frac{1}{4}$
$x^{2} = \frac{1}{2}$
$x = \sqrt{\frac{1}{2}} = \frac{1}{2} \sqrt{2}$ of $x = ‐ \frac{1}{2} \sqrt{2}$

${(\frac{x}{\sqrt{2}})}^{2} = \frac{1}{4}$

tweede manier
$\frac{x}{\sqrt{2}} = \frac{1}{2}$ of $\frac{x}{\sqrt{2}} = ‐ \frac{1}{2}$
$x = \frac{1}{2} \sqrt{2}$ of $x = ‐ \frac{1}{2} \sqrt{2}$

${(\frac{x}{\sqrt{2}})}^{2} = 8$

eerste manier
$\frac{x^{2}}{2} = 8$
$x^{2} = 16$
$x = 4$ of $x = ‐ 4$

${(\frac{x}{\sqrt{2}})}^{2} = 8$

tweede manier
$\frac{x}{\sqrt{2}} = \sqrt{8}$ of $\frac{x}{\sqrt{2}} = ‐ \sqrt{8}$
$x = \sqrt{8} \cdot \sqrt{2} = \sqrt{16} = 4$ of $x = ‐ 4$

$v = 11,5 \sqrt{25,6} \approx 58,2$ km/u

$100 = 11,5 \sqrt{r}$
$\frac{100}{11,5} = \sqrt{r}$
${(\frac{100}{11,5})}^{2} = r$
$r \approx 75,6$ meter

Een kubus heeft zes vierkante grensvlakken van 2 bij 2 de oppervlakte van elk grensvlak is 4, dus $6 \cdot 4 = 24$ .

$O = 6 r^{2}$

$r = \sqrt{\frac{11}{6}} = \frac{1}{6} \sqrt{66}$

$r = \sqrt{\frac{O}{6}} = \sqrt{\frac{6 O}{36}} = \sqrt{\frac{1}{36}} \cdot \sqrt{6 O} = \frac{1}{6} \sqrt{6 O}$

$2 \cdot 2 + \sqrt{12} = 4 + 2 \sqrt{3}$

$2 \cdot (\sqrt{2} + 2 \sqrt{2} + \sqrt{6}) = 6 \sqrt{2} + 2 \sqrt{6}$

$2 \cdot 3 - \sqrt{18} = 6 - 3 \sqrt{2}$	$4 \sqrt{18} - 2 \cdot 3 = 12 \sqrt{2} - 6$
$2 \cdot \sqrt{1} - \sqrt{3} = 2 - \sqrt{3}$	$\sqrt{3} + \sqrt{9} = \sqrt{3} + 3$

$\sqrt{5} + \sqrt{3} bij \sqrt{5} - \sqrt{3}$

$5 - 3 = 2$

$(\sqrt{5} + \sqrt{3}) \cdot (\sqrt{5} - \sqrt{3}) = 5 - 3 = 2$

$4$ ; $3$ en $2$ keer $2 \sqrt{3}$

$7 + 4 \sqrt{3}$

$k (a + b) = k a + k b$	$k (a - b) = k a - k b$
${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$	${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
$(a + b) (a - b) = a^{2} - b^{2}$	$(a + b) (c + d) = a c + b c + a d + b d$

Linkerkolom:
$3 + 2 \cdot \sqrt{3} \cdot 1 + 1 = 4 + 2 \sqrt{3}$
$3 + 2 \cdot \sqrt{3} \cdot \sqrt{2} + 2 = 5 + 2 \sqrt{6}$
$3 - \sqrt{3}$
$3 - 1 = 2$

Rechterkolom:
$3 - 2 \cdot \sqrt{3} \cdot 1 + 1 = 4 - 2 \sqrt{3}$
$3 - 2 \cdot \sqrt{3} \cdot \sqrt{2} + 2 = 5 - 2 \sqrt{6}$
$3 - \sqrt{6}$
$\sqrt{12} - \sqrt{6} + \sqrt{6} - \sqrt{3} = 2 \sqrt{3} - \sqrt{3} = \sqrt{3}$