Question

efectuati: VA ROG sa ma ajutati!

Answer 1

 

$\displaystyle\bf\\6^*.\\a)\\\Big(5^{17}\cdot5^{18}+7^{23}:7^{15}\Big):\Big(7\cdot7^2\cdot7^5+5^{50}:5^{15}\Big)=\\\\=\Big(5^{17+18}+7^{23-15}\Big):\Big(7^{1+2+5}+5^{50-15}\Big)=\\\\=\Big(5^{35}+7^{8}\Big):\Big(7^{8}+5^{35}\Big)=\\\\=\Big(5^{35}+7^{8}\Big):\Big(5^{35}+7^{8}\Big)=\boxed{\bf1}$

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$\displaystyle\bf\\b)\\\left[\Big(2^{10}\Big)^8+6^5\cdot3^7+2^{3^2}\right]:\left[\Big(2^5\Big)^{16}+6^{12}:2^7+2^9\right]=\\\\\\=\left[\Big(2^{10\times8}\Big)+(2\times3)^5\cdot3^7+2^9\right]:\left[\Big(2^{5\times16}\Big)+(2\times3)^{12}:2^7+2^9\right]=\\\\\\=\Big[2^{80}+2^5\cdot3^5\cdot3^7+2^9\Big]:\Big[2^{80}+2^{12}\cdot3^{12}:2^7+2^9\Big]=$

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$\displaystyle\bf\\=\Big[2^{80}+2^5\cdot3^{5+7}+2^9\Big]:\Big[2^{80}+2^{12-7}\cdot3^{12}+2^9\Big]=\\\\\\=\Big[2^{80}+2^5\cdot3^{12}+2^9\Big]:\Big[2^{80}+2^5\cdot3^{12}+2^9\Big]=\boxed{\bf1}$