Răspuns:
Utilizăm regulile de calcul cu puteri.
La înmulțire adunăm exponenții, iar la împărțire îi scădem
[tex]a) \ -{2}^{5} \cdot ( {2}^{3} \cdot {2}^{5} )^{2} :(-4)^{9} = -{2}^{5} \cdot ({2}^{3 + 5})^{2} :(- {2}^{2} )^{9} = -{2}^{5} \cdot ({2}^{8})^{2} :(- {2}^{2 \cdot 9} )= -{2}^{5} \cdot {2}^{8 \cdot 2} :(- {2}^{18} ) = {2}^{5} \cdot {2}^{16} :{2}^{18} = {2}^{5 + 16 - 18} = {2}^{3} = \bf 8[/tex]
[tex]b) - {3}^{5} \cdot ( {3}^{4} \cdot {3}^{6} )^{2} :(-27)^{8} = - {3}^{5} \cdot ( {3}^{4 + 6})^{2} : 27^{8} = - {3}^{5} \cdot ( {3}^{10})^{2} : ( {3}^{3} )^{8} = - {3}^{5} \cdot {3}^{10 \cdot 2} : {3}^{3 \cdot8} = - {3}^{5} \cdot {3}^{20} : {3}^{24} = - {3}^{5 + 20 - 24} = -{3}^{1} = \bf - 3[/tex]
[tex]c) [(-2)^{6} \cdot 2^{2} \cdot 2^{4}]:[(-2)^{2} \cdot 2^{3} \cdot (-2)^{4}]:(-4) = (2^{6} \cdot 2^{2} \cdot 2^{4}):(2^{2} \cdot 2^{3} \cdot 2^{4}):(-4) = 2^{6 + 2 + 4}:2^{2 + 3 + 4}:(-4) = 2^{12}:2^{9}:(-4) = 2^{12 - 9}:(-4) = 2^{3}:(-4) = 8:(-4) = \bf - 2[/tex]
[tex]d) [(-3)^{4} \cdot {3}^{3} \cdot (-3)^{6}]:[3^{3} \cdot (-3)^{2}\cdot (-3)^{6}]:(-9) = (3^{4} \cdot {3}^{3} \cdot 3^{6}):(3^{3} \cdot 3^{2}\cdot 3^{6}):(-9) = {3}^{4 + 3 + 6} : 3^{3 + 2 + 6} : (-9) = {3}^{13} : 3^{11} : (-9) = {3}^{13 - 11} : (-9) = {3}^{2} : (-9) = 9 : (-9) = \bf - 1[/tex]
[tex]e) [(-2)^{2} \cdot 2^{3} \cdot (-2)^{6}]^{2} : 4^{8} = (2^{2} \cdot 2^{3} \cdot 2^{6})^{2} : ( {2}^{2} )^{8} = (2^{2 + 3 + 6})^{2} : {2}^{2 \cdot 8} = (2^{11})^{2} : {2}^{16} = 2^{22} : {2}^{16} = 2^{22 - 16} = 2^{6} = \bf 64[/tex]
[tex]f) [(-3)^{2} \cdot 3^{3} \cdot (-3)^{4}]^{2} : 27^{5} = (3^{2} \cdot 3^{3} \cdot 3^{4})^{2} : ( {3}^{3} )^{5} = (3^{2 + 3 + 4})^{2} : {3}^{3 \times 5} = (3^{9})^{2} : {3}^{15} = 3^{9 \cdot 2} : {3}^{15} = 3^{18} : {3}^{15} = 3^{18 - 15} = 3^{3} = \bf 27[/tex]
