Math, asked by tusharghosh, 1 year ago

evaluate all of these​

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Answered by ananya9703
6

I) 3^3 × (243)^-2/3 × 9^-1/3

= 3^3 × (3^5)^-2/3 × (3^2)^-1/3

= 3^3 × 3^-10/3 × 3^-2/3

= 3^(3 -10/3 -2/3) ( as a^m × a^n = a^(m + n)

= 3^(3 - 12/3)

= 3^(3 - 4)

= 3^(-1) ( as a^ - m = 1/a^m )

= 1/3

II) 5^ -4 × (125)^5/3 ÷ (25)^ -1/2

= 5^-4 × (5^3)^5/3 ÷ (5^2)^-1/2

= 5^-4 × (5)^3×5/3 ÷ (5)^2×-1/2

= 5^-4 × (5)^5 ÷ (5)^-1 (as a^m × a^n = a^m+n )

= 5^(-4 + 5) ÷ 5^-1 (as a^-m = 1/a^m)

= 5^1 ÷ 1/5

= 5 ÷ 1/5

= 5 × 5

= 25

III) (27/125)^2/3 × (9/25)^-3/2

= (3^3/5^3)^2/3 × (3^2/5^2)^-3/2 ( (a^m/b^m) = (a/b)^m )

= (3/5)^3×2/3 × (3/5)^2×-3/2

= (3/5)^2 × (3/5)^-3

= (3/5)^(2-3) ( a^m × a^n = a^m+n)

= (3/5)^-1 ( a^-m = 1/a^m )

= 5/3

IV) 7^0 × 25^-3/2 - 5^-3

= 1 × (5^2)^-3/2 - 5^-3 ( (a^m)^n = a^m×n )

= (5)^2×-3/2 - (5)^-3

= (5)^-3 - (5)^-3

= 0

V) (16/81)^-3/4 × (49/9)^3/2 ÷ (343/216)^2/3

= (2^4/3^4)^-3/4 × (7^2/3^2)^3/2 ÷ (7^3/6^3)^2/3

= (2/3)^4 × -3/4 × (7/3)^2 × 3/2 ÷ (7/6)^3 × 2/3

= (2/3)^-3 × (7/3)^3 ÷ (7/6)^2

= (3/2)^3 × (7/3)^3 × (6/7)^2

= 3^3/2^3 × 7^3/3^3 × 6^2/7^2

= 9 × 7/2 = 63/2

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