Saidejan
Rozwiązane

Pomocy!!!
Zadanie drugie i trzecie daje naj!!!! ​



Pomocy Zadanie Drugie I Trzecie Daje Naj class=

Odpowiedź :

Matfizianka06

Odpowiedź:

[tex]2. \: a)(1 \frac{1}{4} - \frac{2}{3} {)}^{2} = (1 \frac{3}{12} - \frac{8}{12} {)}^{2} = ( \frac{7}{12} {)}^{2} = \frac{49}{144} \\ b) - 2 \times {5}^{2} = - 2 \times 25 = - 50 \\ c) \frac{ { - 4}^{2} }{ \sqrt{25 - 9} } = \frac{ - 16}{ \sqrt{16} } = \frac{ - 16}{4} = - 4 \\ d)( \sqrt{121} - 7 \frac{1}{2} {)}^{2} = (11 - 7 \frac{1}{2} {)}^{2} = (3 \frac{1}{2} {)}^{2} = ( \frac{7}{2} {)}^{2} = \frac{49}{4} = 12 \frac{1}{4} \\ 3.a) \sqrt{ \sqrt{81} } = \sqrt{9} = 3 \\ b) \sqrt[3]{ \sqrt{64} } = \sqrt[3]{8} = 2 \\ c) \sqrt[3]{ \sqrt[3]{1} } = \sqrt[3]{1} = 1 \\ d) \sqrt{ \sqrt[3]{729} } = \sqrt{9} = 3[/tex]

Basetla

2.

[tex]a) \ (1\frac{1}{4}-\frac{2}{3})^{2} = (\frac{5}{4}-\frac{2}{3})^{2} = (\frac{15}{12}-\frac{8}{12})^{2} = (\frac{7}{12})^{2} = \frac{49}{144}\\\\b) \ -2\cdot5^{2} = -2\cdot25 = -50\\\\c) \ \frac{-4^{2}}{\sqrt{25-9}} = \frac{-16}{\sqrt{16}} = \frac{-16}{4} = -4\\\\d) \ (\sqrt{11}-7\frac{1}{2})^{2} =(11-\frac{15}{2})^{2} = (\frac{22}{2}-\frac{15}{2})^{2} = (\frac{7}{2})^{2} = \frac{49}{4} =12\frac{1}{4}[/tex]

3.

[tex]a) \ \sqrt{\sqrt{81}} = \sqrt{9} = 3\\\\b) \ \sqrt[3]{\sqrt{64}} = \sqrt[3]{8} = \sqrt[3]{2^{3}}=2\\\\c) \ \sqrt[3]{\sqrt[3]{1}}} = \sqrt[3]{1} = 1\\\\d) \ \sqrt{\sqrt[3]{729}} = \sqrt{9} = 3[/tex]

Inne Pytanie