Zadanie w załączniku.Jeżeli nie bd można się doczytać proszę do mnie pisać. Z góry dzięęki :D
Jeśli dobrze Cię rozczytałam to będzie tak:
a)[0,6 + (1/3)^-1 * (5 * 3^-1 - sqrt(1/9))]^-1 =
[0,6 + 3*(5*1/3 - 1/3)]^-1 = [0,6 + 3*4/3]^-1 = [0,6 + 4]^-1 = (46/10)^-1 = 10/46
b)[2^-1 * (5^2 - 5)*4]/[2^3 - (-4)^-1 - 2^-4 + 2^5] =
[1/2 * (25 - 5) *4]/[8 - (-1/4) - (1/2)^4 + 32] = (2*20)/[8 + 1/4 - 1/16 + 32] =
40/(128/16 + 4/16 - 1/16 + 512/16) = 40/(643/16) = 40 * 16/643 = 640/643
c)[(1/3)^-4 - 2^4 * 5]/ [(-3)^2 + (1/3)^-2: 3^2 - 3^4: 3^2] =
[3^4 - 16*5]/[9 + 3^2 : 3^2 - 3^2*3^2 : 3^2] = (81- 80)/[3^2 *(1 + 1): 3^2(1 + 3^2) : 3^2]=
1/ [9*2/ 3^2*(1 + 9) / 3^2] = 1/[18/10] = 1 * 10/ 18 = 10/18
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Jeśli dobrze Cię rozczytałam to będzie tak:
a)[0,6 + (1/3)^-1 * (5 * 3^-1 - sqrt(1/9))]^-1 =
[0,6 + 3*(5*1/3 - 1/3)]^-1 = [0,6 + 3*4/3]^-1 = [0,6 + 4]^-1 = (46/10)^-1 = 10/46
b)[2^-1 * (5^2 - 5)*4]/[2^3 - (-4)^-1 - 2^-4 + 2^5] =
[1/2 * (25 - 5) *4]/[8 - (-1/4) - (1/2)^4 + 32] = (2*20)/[8 + 1/4 - 1/16 + 32] =
40/(128/16 + 4/16 - 1/16 + 512/16) = 40/(643/16) = 40 * 16/643 = 640/643
c)[(1/3)^-4 - 2^4 * 5]/ [(-3)^2 + (1/3)^-2: 3^2 - 3^4: 3^2] =
[3^4 - 16*5]/[9 + 3^2 : 3^2 - 3^2*3^2 : 3^2] = (81- 80)/[3^2 *(1 + 1): 3^2(1 + 3^2) : 3^2]=
1/ [9*2/ 3^2*(1 + 9) / 3^2] = 1/[18/10] = 1 * 10/ 18 = 10/18