Odpowiedź:
Poziom A
g)
[tex]3 \sqrt{11} = \sqrt{3 \times 3 \times 11 } = \sqrt{9 \times 11} = \sqrt{99} [/tex]
H)
[tex] \frac{2}{5} \sqrt{15} = \sqrt{ \frac{2}{5} \times \frac{2}{5} \times 15} = \sqrt{ \frac{4}{25} \times 15} = \sqrt{ \frac{60}{25} } [/tex]
Poziom B
g)
[tex] \sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2 \sqrt{6} [/tex]
h)
[tex] \sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5 \sqrt{3} [/tex]
Poziom C
g)
[tex]4 \sqrt{8} - 5 \sqrt{8} + 6 \sqrt{8} = 5 \sqrt{8} [/tex]
h)
[tex] - \sqrt{3} - 3 \sqrt{3} + 10 \sqrt{3} = 6 \sqrt{3} [/tex]
Poziom D
g)
[tex] \sqrt{2} + \sqrt{8} - \sqrt{32} = \sqrt{2} + 2 \sqrt{2} - 4 \sqrt{2} = - \sqrt{2} [/tex]
h)
[tex]3 \sqrt{3} + \sqrt{48} + \sqrt{75} = 3 \sqrt{3} + 4 \sqrt{3} + 5 \sqrt{3} = 12 \sqrt{3} [/tex]
