Question

(2/7)²×(7/3)-³÷{(7/5)-²}-⁴

Answer 1

=(2/7)^2×(7/3)^-3÷{(7/5)^-2}-4

=2^2/7^2×3^3/7^3÷{(7/5)^-2×-4}

=2^2/7^2×3^3/7^3÷{(7/5)^8}

=2^2/7^2×3^3/7^3÷{7^8/5^8}

=2^2/7^2×3^3/7^3×5^8/7^8

=(2^2×3^3×5^8)/(7^5×7^8)

=(2^2×3^3×5^8)/(7^13)

Answer 2

hai friend,

here is your answer,

lets divide the problem into two parts
(2/7)²×(7/3)^-3 =a

{(7/5)^-2}^-4 =b

we have to find a /b

a= (2/7)²*(7/3)^-3

=4/49 *27/343

=2²*3³/7^5


b={(7/5)^-2}^-4
=7/5^8


now

a / b=2²*3³/7^5 /(7/5)^8

=2²*3³*5^8/7^5*7^8

=2²*3³*5^8/7^13

= 2²*3³*5^8*7^-13



hope helped!!