Question

Answer this quetion gast pls if you woll help me i will help you plsssss answer this

Answer 1

Solution

$= > \frac{9 {}^{n} \times 3 {}^{2} \times (3 {}^{ \frac{ - n}{2} } ) {}^{ - 2} - 27 {}^{n} }{3 {}^{3m} \times 2 {}^{3} } = \frac{1}{27} \\ = > \frac{3 {}^{2n} \times 3 {}^{2} \times 3 {}^{n} - 3 {}^{3n} }{3 {}^{3m} \times 2 {}^{3} } = \frac{1}{3 {}^{3} } \\ = > \frac{3 {}^{2n + n} \times 3 {}^{2} - 3 {}^{3n} }{3 {}^{3m} \times 2 {}^{3} } = 3 {}^{ - 3} \\ = > \frac{3 {}^{3n}(3 {}^{2} - 1) }{3 {}^{3m} \times 2 {}^{3} } = 3 {}^{ - 3} \\ = > \frac{3 {}^{3n} \times 8 }{3 {}^{3m} \times 8 } = 3 {}^{ - 3} \\ = > 3 {}^{3n - 3m} = 3 {}^{ - 3} \\ therefore..... \\ = > 3n - 3m = - 3 \\ = > m - n = 1 \\ \\ \\ \\ \: \: \: \: \: \: \large\mathfrak{\underline{\huge\mathcal{\bf{\boxed{\boxed{\large\mathcal{m - n = 1}}}}}}}(proved) \:$

Answer 2

Answer:

question is uncomplete , can be true if m =n+ 1. '