Question

please please please please please please please solve this problem sir



please please please please please don't scame​

Answer 1

Step-by-step explanation:

(3²)^n × 3² ×(3^-n/2)^(-2) -(3³)^n ÷ 3^3m ×2³ = 1/3³

a^m× a^n = a^(m+n)

a^m/a^n =a^(m-n)

(a^-m)^-n =a^mn

3^(2n +2)× 3^(n/2×2) - 3^3n ÷ 3^3m×2³

3^(2n+2+n) / 3^3m × 2³ -3^3n/3^3m×2³

3^(3n+2-3m)/ 2³ - 3^(3n-3m)/2³

3^(3n-3m)(3² -1)/2³

3^(3n-3m) 8/8

=3^3(n-m) = 1/3³ = 3^(-3)

bases are same so powers are equal

3(n-m) =-3

m-n =1

Answer 2

$= > \frac{ {9}^{n} \times {3}^{2} \times { {3}^{ \frac{ - n}{2} } }^{ - 2} - {27}^{n} }{ {3}^{ 3m} \times {2}^{3} } = \frac{1}{27}$

$= > \frac{ {9}^{n} \times 9 \times {3}^{n} - {27}^{n} }{ {27}^{m} \times {2}^{3} } = \frac{1}{27}$

$= > \frac{ {27}^{n} ( {9} - 1) }{ {27}^{m} \times 8} = \frac{1}{27}$

$= > {27}^{n + 1} ( 8) = {27}^{m} \times 8$

$= > {27}^{m - n - 1} = 1$

$= > m - n - 1 = 0$

$= > m - n = 1$

Hence, proved.

Hope this answer helps you ^_^ !