Odpowiedź:
zad. 2
[tex] \sqrt{( - 5 {)}^{2} } = \sqrt{25} = 5[/tex]
P
[tex]( \sqrt[3]{( - 6 {)}^{2} {)}^{3} } = \sqrt{( - 6 {)}^{2} } = \sqrt{36} = 6[/tex]
F
Zad. 3
[tex] \frac{ \sqrt{2} }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{2}{2} = 1[/tex]
[tex]2 \sqrt{3} - 2 \sqrt{3} = 0[/tex]
[tex] \frac{ \sqrt{4} }{4} [/tex]
[tex] \frac{25}{ \sqrt{5} } \times \frac{ \sqrt{5} }{ \sqrt{5} } = \frac{25 \sqrt{5} }{5} = 5 \sqrt{5} [/tex]
Odp. D
zad. 4
[tex] \sqrt[3]{ \sqrt{49} + 1 } = \sqrt[3]{7 + 1} = \sqrt[3]{8} = 2[/tex]
odp. C
zad. 5
[tex] \sqrt{ \frac{9}{25} } \times \sqrt{ \frac{25}{36} } + ( \sqrt{361} {)}^{2} - 4 \times \sqrt{0.49} \times \sqrt{100} = \frac{3}{5} \times \frac{5}{6} + 361 - 4 \times 0.7 \times 10 = \frac{1}{2} + 361 - 28 = 333 \frac{1}{2} [/tex]
zad. 6
√60≈7,4
√160≈12,6
liczby całkowite to np. 1,-1,2,-2 ... itd.
czyli liczby całkowite to :
8,9,10,11,12
zad. 7
V=a×b×c
V=4√3×2√2×√3=8√18=24√2
