1.
[tex]a) \: \: \sqrt{81} + \sqrt{9} = 9 + 3 = 12[/tex]
[tex]b)\: \sqrt{64} - \sqrt{25} = 8 - 5 = 3[/tex]
[tex]c) \: \sqrt{49} + \sqrt{36} - \sqrt{25} = 7 + 6 - 5 = 8[/tex]
[tex]d) \: \sqrt{100} \times \sqrt{16} - \sqrt{121} = 10 \times 4 - 11 = 40 - 11 = 29[/tex]
2.
[tex]a) \: 1 - \frac{1}{7} \times \sqrt{1 \frac{27}{169} } = 1 - \frac{1}{7} \times \sqrt{ \frac{196}{169} } = 1 - \frac{1}{7} \times \frac{14}{13} = 1 - \frac{2}{13} = \frac{11}{13} [/tex]
[tex]b) \: 16 - 16 \times \sqrt{5 \frac{1}{16} } = 16 - 16 \times \sqrt{ \frac{81}{16} } = 16 - 16 \times \frac{9}{4} = 16 - 36 = - 20[/tex]
[tex]c) \: 2 + 2 \times \sqrt{1 \frac{9}{16} } = 2 + 2 \times \sqrt{ \frac{25}{16} } = 2 + 2 \times \frac{5}{4} = 2 + \frac{5}{2} = 2 + 2 \frac{1}{2} = 4 \frac{1}{2} [/tex]
[tex]d) \: 4 + \frac{1}{4} \times \sqrt{5 \frac{11}{49} } = 4 + \frac{1}{4} \times \sqrt{ \frac{256}{49} } = 4 + \frac{1}{4} \times \frac{16}{7} = 4 + \frac{4}{7} = 4 \frac{4}{7} [/tex]
Mam nadzieję, że pomogłam! ❤️
Cześć ;-)
Zadanie 2
[tex]a] \ \sqrt{81}+\sqrt9=\sqrt{9^2}+\sqrt{3^2}=9+3=12\\\\b] \ \sqrt{64}-\sqrt{25}=\sqrt{8^2}-\sqrt{5^2}=8-5=3\\\\c] \ \sqrt{49}+\sqrt{36}-\sqrt{25}=\sqrt{7^2}+\sqrt{6^2}-\sqrt{5^2}=7+6-5=8\\\\d] \ \sqrt{100}\cdot\sqrt{16}-\sqrt{121}=\sqrt{10^2}\cdot\sqrt{4^2}-\sqrt{11^2}=10\cdot4-11=40-11=29[/tex]
Zadanie 3
Od a) do b)
[tex]a] \ 1-\frac{1}{7}\cdot\sqrt{1\frac{27}{169}}=1-\frac{1}{7}\cdot\sqrt{\frac{196}{169}}=1-\frac{1}{7}\cdot\sqrt{\frac{14^2}{13^2}}=1-\frac{1}{7}\cdot\frac{14}{13}=1-\frac{2}{13}=\frac{11}{13}\\\\b] \ 16-16\cdot\sqrt{5\frac{1}{16}}=16-16\cdot\sqrt{\frac{81}{16}}=16-16\cdot\sqrt{\frac{9^2}{4^2}}=16-16\cdot\frac{9}{4}=\\\\=16-4\cdot9=16-36=-20[/tex]
Od c) do d)
[tex]c] \ 2+2\cdot\sqrt{1\frac{9}{16}}=2+2\cdot\sqrt{\frac{25}{16}}=2+2\cdot\sqrt{\frac{5^2}{4^2}}=2+2\cdot\frac{5}{4}=2+2\cdot1,25=2+2,5=4,5\\\\d] \ 4+\frac{1}{4}\cdot\sqrt{5\frac{11}{49}}=4+\frac{1}{4}\cdot\sqrt{\frac{256}{49}}=4+\frac{1}{4}\cdot\sqrt{\frac{16^2}{7^2}}=4+\frac{1}{4}\cdot\frac{16}{7}=4+\frac{4}{7}=4\frac{4}{7}[/tex]
