Odpowiedź:
zad. 10
[tex] \sqrt[3]{ - \frac{1}{27} } + \sqrt{1 \frac{7}{9} } = - \frac{1}{3} + \sqrt{ \frac{16}{9} } = - \frac{1}{3} + \frac{4}{3} = \frac{3}{3} = 1[/tex]
zad. 11
[tex] \frac{2}{3 \sqrt{3} } \times \frac{ 3\sqrt{3} }{3 \sqrt{3} } = \frac{6 \sqrt{3} }{27} = \frac{2 \sqrt{3} }{9} [/tex]
zad. 12
[tex]2 - \frac{x + 2}{3} = \frac{x}{2} \: \: \: \: | \times 6 \\ 12 - 2(x + 2) = 3x \\ 12 - 2x - 4 = 3x \\ - 2x - 3x = 4 - 12 \\ - 5x = - 8 \\ x = \frac{8}{5} [/tex]
(0;2)
