Explicație pas cu pas:
a)
[tex]({2}^{4})^{7} - {2}^{28} = {2}^{4 \times 7} - {2}^{28} = {2}^{28} - {2}^{28} = 0 \\ [/tex]
b)
[tex]({3}^{5})^{4} \div {3}^{19} = {3}^{5 \times 4} \div {3}^{19} = {3}^{20} \div {3}^{19} = {3}^{20 - 19} \\= {3}^{1} = 3 [/tex]
c)
[tex]({5}^{8})^{7} \div ({5}^{53} \times 5) = {5}^{8 \times 7} \div {5}^{53 + 1} = {5}^{56} \div {5}^{54} \\ = {5}^{56 - 54} = {5}^{2} = 25[/tex]
d)
[tex]{( {7}^{7} )}^{7} \div ( {7}^{77} \div {7}^{7 \times 5}) = {7}^{7 \times 7} \div ( {7}^{77} \div {7}^{35}) = {7}^{49} \div {7}^{77 - 35} \\ = {7}^{49} \div {7}^{77 - 35} = {7}^{49} \div {7}^{42} = {7}^{49 - 42} = {7}^{7} [/tex]
e)
[tex]{( {13}^{8} )}^{13} \div {( {13}^{26} )}^{4} = {13}^{8 \times 13} \div {13}^{26 \times 4} = {13}^{104} \div {13}^{104} \\= {13}^{104-104} ={13}^{0} = 1 [/tex]
f)
[tex]{7}^{ {13}^{2} } \div {( {7}^{42} )}^{4} = {7}^{169} \div {7}^{42 \times 4} = {7}^{169} \div{7}^{168} \\ = {7}^{169 - 168} = {7}^{1} = 7 \\ [/tex]
g)
[tex]{2}^{ {7}^{2} } \div {( {2}^{8} )}^{6} = {2}^{49} \div {2}^{8 \times 6} ={2}^{49} \div {2}^{48} \\ = {2}^{49 - 48} = {2}^{1} = 2[/tex]
h)
[tex]{3}^{ {8}^{2} } \div {3}^{ {4}^{3} } = {3}^{64} \div {3}^{64} = {3}^{64 - 64} = {3}^{0} = 1 \\ [/tex]
