Ja, nee

${(\sqrt{7} \cdot \sqrt{9})}^{2} = \sqrt{7} \cdot \sqrt{9} \cdot \sqrt{7} \cdot \sqrt{9} = \sqrt{7} \cdot \sqrt{7} \cdot \sqrt{9} \cdot \sqrt{9} = 7 \cdot 9 = 63$

$55$

${\sqrt{5}}^{2} + {\sqrt{3}}^{2} = 8$ , dus de schuine zijde is $\sqrt{8}$ .

De twee rechthoekszijden samen zijn langer dan de schuine zijde (de kortste verbinding van twee punten is een rechte lijn).

$\sqrt{20}$

$\sqrt{50}$

$\sqrt{75}$

$\sqrt{5 a^{2}}$

$\sqrt{4 a}$

$\sqrt{a^{2} b}$

$\sqrt{8} = \sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2}$

$\sqrt{28} = \sqrt{4} \cdot \sqrt{7} = 2 \sqrt{7}$

$\sqrt{18} = \sqrt{9} \cdot \sqrt{2} = 3 \sqrt{2}$

$\sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4 \sqrt{3}$

$\sqrt{20} = \sqrt{4} \cdot \sqrt{5} = 2 \sqrt{5}$

$\sqrt{60} = \sqrt{4} \cdot \sqrt{15} = 2 \sqrt{15}$

$\sqrt{40} = \sqrt{4} \cdot \sqrt{10} = 2 \sqrt{10}$

$\sqrt{80} = \sqrt{16} \cdot \sqrt{5} = 4 \sqrt{5}$

$\sqrt{\frac{7}{25}} = \sqrt{\frac{1}{25}} \cdot \sqrt{7} = \frac{1}{5} \sqrt{7}$

$\sqrt{\frac{7}{100}} = \sqrt{\frac{1}{100}} \cdot \sqrt{7} = \frac{1}{10} \sqrt{7}$

$\sqrt{\frac{7}{36}} = \sqrt{\frac{1}{36}} \cdot \sqrt{7} = \frac{1}{6} \sqrt{7}$

$\sqrt{\frac{3}{49}} = \sqrt{\frac{1}{49}} \cdot \sqrt{3} = \frac{1}{7} \sqrt{3}$

Bovenste rij:
$\sqrt{72} = \sqrt{36} \cdot \sqrt{2} = 6 \sqrt{2}$
$\sqrt{76} = \sqrt{4} \cdot \sqrt{19} = 2 \sqrt{19}$
$\sqrt{80} = \sqrt{16} \cdot \sqrt{5} = 4 \sqrt{5}$
$\sqrt{84} = \sqrt{4} \cdot \sqrt{21} = 2 \sqrt{21}$

Onderste rij:
$\sqrt{\frac{12}{49}} = \frac{1}{7} \sqrt{12} = \frac{1}{7} \cdot \sqrt{4} \cdot \sqrt{3} = \frac{2}{7} \sqrt{3}$
$\sqrt{\frac{48}{49}} = \frac{1}{7} \sqrt{48} = \frac{1}{7} \cdot \sqrt{16} \cdot \sqrt{3} = \frac{4}{7} \sqrt{3}$
$\sqrt{2 \frac{2}{25}} = \sqrt{\frac{52}{25}} = \frac{1}{5} \sqrt{52} = \frac{1}{5} \cdot \sqrt{4} \cdot \sqrt{13} = \frac{2}{5} \sqrt{13}$
$\sqrt{\frac{10}{b^{2}}} = \frac{1}{b} \sqrt{10}$

$0$

$\sqrt{160} = \sqrt{16} \cdot \sqrt{10} = 4 \sqrt{10}$ , $\sqrt{90} = \sqrt{9} \cdot \sqrt{10} = 3 \sqrt{10}$ , dus $\sqrt{160} - \sqrt{90} - \sqrt{10} = 4 \sqrt{10} - 3 \sqrt{10} - \sqrt{10} = 0$

$\sqrt{9} \cdot \sqrt{3} + \sqrt{25} \cdot \sqrt{3} = 3 \sqrt{3} + 5 \sqrt{3} = 8 \sqrt{3}$
$\sqrt{4} \cdot \sqrt{2} + \sqrt{9} \cdot \sqrt{2} = 2 \sqrt{2} + 3 \sqrt{2} = 5 \sqrt{2}$
$\sqrt{10} + \sqrt{100} \cdot \sqrt{10} = \sqrt{10} + 10 \sqrt{10} = 11 \sqrt{10}$

$\sqrt{9} \cdot \sqrt{5} - \sqrt{4} \cdot \sqrt{5} = 3 \sqrt{5} - 2 \sqrt{5} = \sqrt{5}$
$\sqrt{16} \cdot \sqrt{2} - \sqrt{4} \cdot \sqrt{2} = 4 \sqrt{2} - 2 \sqrt{2} = 2 \sqrt{2}$
$\sqrt{100} \cdot \sqrt{3} - \sqrt{3} = 10 \sqrt{3} - \sqrt{3} = 9 \sqrt{3}$

$\sqrt{a} + \sqrt{4} \cdot \sqrt{a} = \sqrt{a} + 2 \sqrt{a} = 3 \sqrt{a}$
$\sqrt{4} \cdot \sqrt{3} + \sqrt{\frac{1}{4}} \cdot \sqrt{3} = 2 \sqrt{3} + \frac{1}{2} \sqrt{3} = 2 \frac{1}{2} \sqrt{3}$
$\sqrt{\frac{1}{4}} \cdot \sqrt{3} + \sqrt{\frac{1}{16}} \cdot \sqrt{3} = \frac{1}{2} \sqrt{3} + \frac{1}{4} \sqrt{3} = \frac{3}{4} \sqrt{3}$

$6$

$\sqrt{12} \cdot \sqrt{3} = \sqrt{36} = 6$

$\sqrt{6} \cdot \sqrt{2} = \sqrt{12} = \sqrt{4} \cdot \sqrt{3} = 2 \sqrt{3}$

$\sqrt{\frac{2}{9}} + \sqrt{\frac{8}{9}} = \frac{1}{3} \sqrt{2} + \frac{2}{3} \sqrt{2} = \sqrt{2}$ ;
$\sqrt{1 \frac{1}{4}} + \sqrt{11 \frac{1}{4}} = \sqrt{\frac{5}{4}} + \sqrt{\frac{45}{4}} = \frac{1}{2} \sqrt{5} + 1 \frac{1}{2} \sqrt{5} = 2 \sqrt{5}$

$2 \sqrt{2} + 4 \sqrt{2} + 8 \sqrt{2} + 16 \sqrt{2} = 30 \sqrt{2}$ , dus het getal $30$ .

$\sqrt{1} = 1$	$\sqrt{16} = 4$	$\sqrt{4} = 2$
$\sqrt{81} = 9$	$\sqrt{10}$	$\sqrt{64} = 8$

$\sqrt{2} + \sqrt{\frac{1}{4}} > \sqrt{2 \frac{1}{4}}$	$\sqrt{9} + \sqrt{16} > \sqrt{25}$
$4 \neq 2$	$\sqrt{1} + \sqrt{100} > \sqrt{101}$

10s

12s

$3 + 4 = 7 = \sqrt{x}$ , dus $x = 7^{2} = 49$

$10 + 0,1 = 10,1 = \sqrt{x}$ , dus $x = {10,1}^{2} = 102,01$

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

$x \sqrt{0,02} = \sqrt{2}$ , dus $x = \frac{\sqrt{2}}{\sqrt{0,02}} = \sqrt{100} = 10$

$\sqrt{x} = 10 \sqrt{3} - 2 \sqrt{3} = 8 \sqrt{3}$ , dus $x = {(8 \sqrt{3})}^{2} = 64 \cdot 3 = 192$

$1 = (\sqrt{x} + 1) (\sqrt{x} - 1) = x - 1$ , dus $x = 2$