Question

coroana si 50 de puncte​

Answer 1

Explicație pas cu pas:

16.a)

${( {25}^{4} )}^{3} \div ( {4}^{2} + {3}^{2})^{11} - ( {0}^{2010} + {1}^{2012} + {2013}^{0}) = {25}^{12} \div {25}^{11} - (0 + 1 + 1) = {25}^{12 - 11} - 2 = 25 - 2 = 23$

b)

$(123 + {2}^{7} \times {2}^{2} + {2}^{67} \div {2}^{27}) \div (123 + {2}^{14} \div {2}^{5} + {( {2}^{5} )}^{8}) = (123 + {2}^{7 + 2} + {2}^{67 - 27}) \div (123 + {2}^{14 - 5} + {2}^{5 \times 8}) = (123 + {2}^{9} + {2}^{40} ) \div (123 + {2}^{9} + {2}^{40} ) = 1$

c)

$({2}^{8} + {9}^{2} + {5}^{6} + {36}^{2}) \div ( {0}^{6} + {4}^{4} + {3}^{4} + {25}^{3} + {6}^{4}) = ({2}^{8} + {( {3}^{2}) }^{2} + {5}^{6} + {( {6}^{2} )}^{2}) \div (0 + {( {2}^{2} )}^{4} + {3}^{4} + {( {5}^{2} )}^{3} + {6}^{4}) = ({2}^{8} + {3}^{4} + {5}^{6} + {6}^{4}) \div({2}^{8} + {3}^{4} + {5}^{6} + {6}^{4}) = 1$