Daje dużo punktów. Bardzo proszę o pomoc.... Question from @llaurensay

Ola9999pl

Odpowiedź:

zadania C nie wiem jak rozwiązać ;(

Szczegółowe wyjaśnienie:

zadania w załączniku ^^

0 votes Thanks 0

basetla

$a) \ 1,5+0,5:(-1\frac{1}{3})+\frac{1}{3} =\frac{15}{10}+\frac{5}{10}:(-\frac{4}{3})+\frac{1}{3}=\frac{3}{2}+\frac{1}{2}\cdot(-\frac{3}{4})+\frac{1}{3} =\frac{3}{2}-\frac{3}{8}+\frac{1}{3}=\\\\=\frac{36}{24}-\frac{9}{24}+\frac{8}{24} = \frac{35}{24} = 1\frac{11}{24}\\\\\\b) \ \sqrt{10^{2}-8^{2}} = \sqrt{100-64} = \sqrt{36} = \sqrt{6^{2}} = 6$

$c) \ (-8)^{-\frac{1}{3}}+(\frac{2}{3})^{-1} =(-\frac{1}{8})^{\frac{1}{3}}+(\frac{3}{2})^{1}} = ((-\frac{1}{2})^{3})^{\frac{1}{3}}+\frac{3}{2} = (-\frac{1}{2})^{1}+\frac{3}{2} =\\\\= -\frac{1}{2}+\frac{3}{2} = \frac{2}{2} = 1$

$d) \ \sqrt{48}-(2\sqrt{3})^{3} =\sqrt{16\cdot3}-2^{3}\cdot\sqrt{3}^{3}=\sqrt{16}\cdot\sqrt{3}-8\cdot3\cdot\sqrt{3} = 4\sqrt{3}-24\sqrt{3} = -20\sqrt{3}$

$e) \ log_{3}27} = log_{3}3^{3} = 3log_{3}3 = 3$

$f) \ log_{2}\frac{1}{48}+log_{2}3 = log_{2}(\frac{1}{48}\cdot3) = log_{2}\frac{1}{16} = log_{2}16^{-1} = log_{2}(2^{4})^{-1} =log_{2}2^{-4} =\\\\=-4log_{2}2 = -4$

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