Odpowiedź:
Poziom d
a)
[tex]4 \sqrt{3} - \sqrt{27} - 2 \sqrt{12} = 4 \sqrt{3} - 3 \sqrt{3} - 4 \sqrt{3} = - 3 \sqrt{3} [/tex]
b)
[tex]2 \sqrt{125} - 3 \sqrt{45} + 4 \sqrt{5} = 2 \times 5 \sqrt{5} - 3 \times 3 \sqrt{5} + 4 \sqrt{5} = 10 \sqrt{5} - 9 \sqrt{5} + 4 \sqrt{5} = 5 \sqrt{5} [/tex]
c)
[tex] \sqrt{200} - \sqrt{162} + \sqrt{2} = 10 \sqrt{2} - 9 \sqrt{2} + \sqrt{2} = 2 \sqrt{2} [/tex]
2
a)
[tex] \sqrt{2} \times \sqrt{8} = \sqrt{2 \times 8} = \sqrt{16} = 4[/tex]
c)
[tex] \sqrt{225 - 144} = \sqrt{81} = 9[/tex]
g)
[tex] \sqrt{64 \times 25} = \sqrt{64} \times \sqrt{25} = 8 \times 5 = 40[/tex]
