Question

(FATEC-SP) Se A= 0,666..., B= 1,333... e C= 0,141414..., então AB-¹ + C é igual a: *



-74/99

127/198

80/99

187/30

67/30

Answer 1

A = 0,666..... = 6/9 = 2/3 ***
B =1,333..... = ( 13 - 1)9 = 12/9 = 4/3 ***
C =0,1414... = 14/99 ***

AB^-1  + C =
[( 2/3).( 4/3)^-1 ] + 14/99 =
[ ( 2/3) . ( 3/4)¹ ] + 14/99 =
( 6/12  ou  1/2 ) + 14/99
1/2 + 14/99 =
mmc = 198
( 99 + 28 )/198 = 127/198  ( b )