Question

es esta amigo(a) ya te lo tome​

Answer 1

Explicación paso a paso:

12)

$\frac{\left(7^5\right)^2}{7^9}+\frac{\left(11^2\right)^3}{11^5}+\frac{\left(8^3\right)^2}{4^9}\\\\\frac{7^{10}}{7^9}+\frac{11^6}{11^5}+\frac{8^6}{4^9}\\\\7^{10-9}+11^{6-5}+\frac{\left(2^3\right)^6}{(2^{2})^{9}  } \\\\7+11+\frac{2^{18}}{2^{18}}\\\\7+11+1\\\\=19$

13)

$9^{-2^{-1}}\\\\\frac{1}{9^{2^{-1}}}\\\\\frac{1}{9^{\frac{1}{2} } } \\\\\frac{1}{\sqrt{9} } \\\\=\frac{1}{3}$

14)

I.

$3^2\cdot \:4^2=\left(3+4\right)^2\\\\9\cdot \:16=7^2\\\\144=49$

II.

$8^{-2}=\frac{2}{8}\\\\\frac{1}{8^2}=\frac{1}{4}\\\\\frac{1}{64}=\frac{1}{4}$

III.

$\sqrt[5]{8} \sqrt[5]{4}=2\\\\\sqrt[5]{2^3}\sqrt[5]{2^2}=2\\\\2^{\frac{3}{5}}\cdot \:2^{\frac{2}{5}}=2\\\\2^{\frac{3}{5}+\frac{2}{5}}=2\\\\2^1=2\\\\2=2$

El resultado es: FFV

15)

$\frac{18^4\cdot \:20}{72^2\cdot \:15^3}\\\\\frac{\left(3^2\cdot \:2\right)^4.2^2\cdot \:5}{\left(2^3\cdot \:3^2\right)^2.\left(3\cdot \:5\right)^3} \\\\\frac{2^4\left(3^2\right)^4.2^2\cdot \:5}{\left(2^3\right)^2\left(3^2\right)^2.3^3\cdot \:5^3} \\\\\frac{3^8\cdot \:2^4.2^2\cdot \:5}{2^6\cdot \:3^4.3^3\cdot \:5^3} \\\\\frac{3^8\cdot \:2^6\cdot \:5}{3^7\cdot \:2^6\cdot \:5^3}\\\\\frac{2^6\cdot \:5\cdot \:3^{8-7}}{2^6\cdot \:5^3}\\\\\frac{2^6\cdot \:3\cdot \:5}{2^6\cdot \:5^3}\\\\[tex]\frac{3\cdot \:5}{5^3}\\\\\frac{3}{5^2}\\\\=\frac{3}{25}$

16)          

$\frac{4^m\cdot \:4^3\cdot \:4^6}{2^{2m}\cdot \:4^5}\\\\\frac{4^{m+3+6}}{2^{2m}.(2^{2})^{5}   } \\\\\frac{4^{m+9}}{2^{2m}.2^{10} } \\\\\frac{4^{m+9}}{2^{2m+10} } \\\\\frac{(2^{2})^{m+9}  }{2^{2m+10} }\\\\\frac{2^{2m+18} }{2^{2m+10} } \\\\2^{2m+18-\left(2m+10\right)}\\\\2^8\\\\=256$

17)

$\frac{\sqrt[3]{\sqrt{\sqrt[3]{4^9}}}}{64^{\frac{2}{3}}}\\\\\frac{2}{2^4}\\\\\frac{1}{2^3}\\\\=\frac{1}{8}$

18)

$\frac{\left(\sqrt[5]{\sqrt{7}}\right)^{20}\left(\sqrt[3]{\sqrt{3}}\right)^{18}}{21}\\\\\frac{7^2\left(\sqrt[3]{\sqrt{3}}\right)^{18}}{21}\\\\\frac{3^3\cdot \:7^2}{21}\\\\\frac{3^3\cdot \:7^2}{3\cdot \:7}\\\\3^2\cdot \:7\\\\9\cdot \:7\\\\=63$

19) No la veo bien