Prosze o odpowiedz z rozwiazaniem (:
(1 1/7)^2 + 4/7^3 - (5/8)^0
(1 1/7)^2 + (4/7)^3 - ( 5/8)^0 = (8/7) ^2 + 64/343 -1 = 64/49 +64/343 - 1 = 448/343 + 64/343 - 1 = 512/343-1= 1 169/343 - 1 = 169/343
( (5/8)^0 = 1 ) :)
(1 i1/7 )^2 + (4/7)^3 - (5/8)^0 = (8/7)^2 + (4/7)^3 -(5/8)^0 = 64/49 + 64/343 -1 =
= 448/343 +64/343 -1 = 512/343 -1 = 1 i 169/343 - 1 = 169/343
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(1 1/7)^2 + (4/7)^3 - ( 5/8)^0 = (8/7) ^2 + 64/343 -1 = 64/49 +64/343 - 1 = 448/343 + 64/343 - 1 = 512/343-1= 1 169/343 - 1 = 169/343
( (5/8)^0 = 1 ) :)
(1 i1/7 )^2 + (4/7)^3 - (5/8)^0 = (8/7)^2 + (4/7)^3 -(5/8)^0 = 64/49 + 64/343 -1 =
= 448/343 +64/343 -1 = 512/343 -1 = 1 i 169/343 - 1 = 169/343