Răspuns:
[tex] ({2}^{3} ) ^{15} \div {4}^{22} + {18}^{5} \div {( {3}^{2} + 2) }^{4} + {12}^{4} \div {( {2}^{2} \times 3 })^{4} \\ = {2}^{45} \div ( {2}^{2} ) ^{22} + {18}^{5} \div {11}^{4} + {12}^{4} \div {12}^{4} \\ = {2}^{45} \div {2}^{44} + {18}^{5} \div {11}^{4} + 1 \\ = 2 + 1 + {18}^{5} \div {11}^{4} = 3 + {18}^{5} \div {11}^{4} [/tex]
Dacă acolo ar fi fost 11 la a 5 a dădea 11 +3=14
