Explicație pas cu pas:
a)
[tex] {1999}^{2} - 1999 \cdot 1998 = 1999(1999 - 1998) = 1999 \cdot 1 = \bf 1999[/tex]
b)
[tex]{3}^{80} - 2 \cdot {3}^{79} - 2 \cdot {3}^{78} = {3}^{78} \cdot ( {3}^{2} - 2 \cdot 3 - 2) = {3}^{78} \cdot (9 - 6 - 2) = {3}^{78} \cdot 1 = \bf {3}^{78}[/tex]
c)
[tex]{2}^{10} - {2}^{9} - {2}^{8} - {2}^{7} - {2}^{6} - ... - 2 = {2}^{9} \cdot (2 - 1) - {2}^{8} - {2}^{7} - {2}^{6} - ... - 2 = {2}^{9} \cdot 1 - {2}^{8} - {2}^{7} - {2}^{6} - ... - 2 = {2}^{9} - {2}^{8} - {2}^{7} - {2}^{6} - ... - 2 = {2}^{8} \cdot (2 - 1) - {2}^{7} - {2}^{6} - ... - 2 = {2}^{8} - {2}^{7} - {2}^{6} - ... - 2 = {2}^{7} \cdot (2 - 1) - {2}^{6} - ... - 2 = {2}^{7} - {2}^{6} - ... - 2 = ... = {2}^{2} - 2 = 2 \cdot (2 - 1) = \bf 2[/tex]
d)
[tex]{2}^{3n + 1} + {4}^{n + 1} \cdot {2}^{n + 2} - {8}^{n + 1} = {2}^{3n + 1} + {( {2}^{2} )}^{n + 1} \cdot {2}^{n + 2} - {( {2}^{3} )}^{n + 1} = {2}^{3n + 1} + {2}^{2n + 2} \cdot {2}^{n + 2} - {2}^{3n + 3} = {2}^{3n + 1} + {2}^{3n + 4} - {2}^{3n + 3} = {2}^{3n + 1} \cdot (1 + {2}^{3} - {2}^{2}) = {2}^{3n + 1} \cdot (1 + 8 - 4) = \bf 5 \cdot {2}^{3n + 1}[/tex]
e)
[tex]{4}^{n + 1} - {2}^{2n} = {( {2}^{2} )}^{n + 1} - {2}^{2n} = {2}^{2n + 2} - {2}^{2n} = {2}^{2n} \cdot ( {2}^{2} - 1) = {2}^{2n} \cdot (4 - 1) = \bf 3 \cdot {2}^{2n}[/tex]
f)
[tex]9 \cdot {25}^{n} - {5}^{2n} = 9 \cdot {({5}^{2})}^{n} - {5}^{2n} = 9 \cdot {5}^{2n} - {5}^{2n} = {5}^{2n} \cdot (9 - 1) = \bf 8 \cdot {5}^{2n}[/tex]
g)
[tex]{4}^{5} - {4}^{4} \cdot 3 - {4}^{3} \cdot 3 - {4}^{2} \cdot 3 - 4 \cdot 3 = {4}^{4} \cdot (4 - 3) - {4}^{3} \cdot 3 - {4}^{2} \cdot 3 - 4 \cdot 3 = {4}^{4} \cdot 1 - {4}^{3} \cdot 3 - {4}^{2} \cdot 3 - 4 \cdot 3 = {4}^{4} - {4}^{3} \cdot 3 - {4}^{2} \cdot 3 - 4 \cdot 3 = {4}^{3} \cdot (4 - 3) - {4}^{2} \cdot 3 - 4 \cdot 3 = {4}^{3} \cdot 1 - {4}^{2} \cdot 3 - 4 \cdot 3 = {4}^{3} - {4}^{2} \cdot 3 - 4 \cdot 3 = {4}^{2} \cdot (4 - 3) - 4 \cdot 3 = {4}^{2} \cdot 1 - 4 \cdot 3 = {4}^{2} - 4 \cdot 3 = 4 \cdot (4 - 3) = 4 \cdot 1 = \bf 4[/tex]
