Odpowiedź:
Szczegółowe wyjaśnienie:
Stosowane tu będą działania na potęgach ujemnych:
a^-n = 1/a^n, np., 3^-2 = 1/3² = 1/9
(a/b)^-n = (b/a)^n, np., (2/3)^{-2} = (3/2)²
(2/3)^{-2} = (3/2)² = 9/4 = 8/4 + 1/4 = 2 1/4 (dwie całe i 1/4 = 2¹/₄)
(2,8)^{-4) * (5/14)^{-4} = (28/10)^{-4} * (14/5)⁴ = (10/28)⁴ * (14/5)⁴ =
= (5/14 * 14/5)⁴ = (5*14/14*5)⁴ = 1⁴ = 1
(5 1/3)³ * (0,75)^{-3} = (15/3 + 1/3)³ * (75/100)^{-3} = (16/3)³ * (3/4)^{-3} =
(16/3)³ * (4/3)³ = (16*4/3*3)³ = (64/9)³ = 262144/729 =
[359*729/729 = 261711/729 = 359] to
= 261711/729 + 433/729 =
= 359 + 433/729 = 359 433/729 (359 całych i 433/729)
(1 2/3)^10 * (1 2/3)^{-8} = (5/3)^10 * (5/3)^{-8} = (5/3)^10 * (3/5)^8 =
[przy mnożeniu liczb potęgowanych wykładniki potęg dodajemy to
(5/3)^2 * [(5/3)^8 = (5/3)^{2 + 8 = 10} = (5/3)^10]
= (5/3)^2 * [(5/3)^8 * (3/5)^8] = (5/3)^2 * 1 = (5/3)^2 = (5/3)² =
= 25/9 = 18/9 + 7/9 = 2 7/9 (2 całe i 7/9)
(1 5/11)^{-2} * (4/11)^{- 2} = (11/11 + 5/12)^{-2} * (4/11)^{- 2} =
= (16/11)^{-2} * (4/11)^{- 2} = (16*4/11*11) ^{- 2} = (64/121)^{- 2} =
(121/64)^{2} = 14641/4096 =
[3*4096/4096 = 12288/4096] to
= 12288/4096 + 2353/4096 = 3 + 2353/4096 (3 całe i 2353/4096)
