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9. Przedstaw w postaci jednej potęgi. ​



9 Przedstaw W Postaci Jednej Potęgi class=

Odpowiedź :

Magda

Odpowiedź:

[tex]a)\ \ 9\cdot3^{11}=3^2\cdot3^{11}=3^{2+11}=3^{13}\\\\\\b)\ \ 2^{13}:32=2^{13}:2^5=2^{13-5}=2^8\\\\\\c)\ \ 2^{50}:(16\cdot2^{10})=2^{50}:(2^4\cdot2^1^0)=2^{50}:2^{4+10}=2^{50}:2^{14}=2^{50-14}=2^{36}\\\\\\d)\ \ \dfrac{125\cdot5^6}{5^7}=\dfrac{5^3\cdot5^6}{5^7}=\dfrac{5^{3+6}}{5^7}=\dfrac{5^9}{5^7}=5^{9-7}=5^2\\\\\\e)\ \ \dfrac{5^2^0}{5^5\cdot5^1^2\cdot25}=\dfrac{5^2^0}{5^1^7\cdot5^2}=\dfrac{5^2^0}{5^1^9}=5^{20-19}=5^1[/tex]

[tex]f)\ \ \dfrac{20^1^0}{20\cdot20^6\cdot400}=\dfrac{20^1^0}{20^1\cdot20^6\cdot20^2}=\dfrac{20^1^0}{20^{1+6+2}}=\dfrac{20^1^0}{20^9}=20^{10-9}=20^1\\\\\\g)\ \ \dfrac{4\cdot2^7}{8\cdot2^5}=\dfrac{2^2\cdot2^7}{2^3\cdot2^5}=\dfrac{2^{2+7}}{2^{3+5}}=\dfrac{2^9}{2^8}=2^{9-8}=2^1\\\\\\h)\ \ (81\cdot3^1^5):(3^7\cdot27)=(3^4\cdot3^1^5):(3^7\cdot3^3)=3^{4+15}:3^{7+3}=3^{19}:3^{10}=\\\\=3^{19-10}=3^9\\\\\\Wykorzystano\ \ wlasno\'sci\ \ poteg\\\\a^{m}\cdot a^{n}=a^{m+n}\\\\a^{m}:a^{n}=a^{m-n}[/tex]

Agulka1987

Odpowiedź:

[tex]a) \: \: 9 \times {3}^{11} = {3}^{2} \times {3}^{11} = {3}^{2 + 11} = {3}^{13} [/tex]

[tex]b) \: \: {2}^{13} \div 32 = {2}^{13} \div {2}^{5} = {2}^{13 - 5} = {2}^{8} [/tex]

[tex] c) \: \: {2}^{50} \div (16 \times {2}^{10} ) = {2}^{50} \div ( {2}^{4} \times {2}^{10} ) = {2}^{50} \div {2}^{4 + 10} = {2}^{50} \div {2}^{14} = {2}^{50 - 14} = {2}^{36} [/tex]

[tex]d) \: \: \frac{125 \times {5}^{6} }{? {5}^{7} } = \frac{ {5}^{3} \times {5}^{6} }{ {5}^{7} } = \frac{ {5}^{9} }{ {5}^{7} } = {5}^{9 - 7} = {5}^{2} [/tex]

[tex]e) \: \: \frac{ {5}^{20} }{ {5}^{5} \times {5}^{12} \times 25} = \frac{ {5}^{20} }{ {5}^{5 + 12} \times {5}^{2} } = \frac{ {5}^{20} }{ {5}^{19} } = {5}^{20 - 19} = {5}^{1} [/tex]

[tex]f) \: \: \frac{ {20}^{10} }{20 \times {20}^{6} \times 400} = \frac{ {20}^{10} }{ {20}^{7} \times {20}^{2} } = \frac{ {20}^{10} }{ {20}^{9} } = {20}^{1} [/tex]

[tex]g) \: \: \frac{4 \times {2}^{7} }{8 \times {2}^{5} } = \frac{ {2}^{2} \times {2}^{7} }{ {2}^{3} \times {2}^{5} } = \frac{ {2}^{9} }{ {2}^{8} } = {2}^{1} [/tex]

[tex]h) \: \: (81 \times {3}^{15} ) \div ( {3}^{7} \times 27) = ( {3}^{4} \times {3}^{15} ) \div ( {3}^{7} \times {3}^{3} ) = {3}^{19} \div {3}^{10} = {3}^{9} [/tex]

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