Question

show that (a^m÷a^-n)^m-n×(a^n÷a^-1)^n-1×(a^1÷a^-m)^1-m =1
​

Answer 1

Answer:

$( \frac{a {}^{m} }{a {}^{ - n} } ) {}^{m - n} \times (\frac{a {}^{n} }{a {}^{ - 1} } ) {}^{n - 1} \times (\frac{a {}^{1} }{a {}^{ - m} } ) {}^{1 - m} \\ \\ =( a {}^{m - ( - n)} ) {}^{m - n} \times (a {}^{n - ( - 1)} ) {}^{n - 1} \times (a {}^{1 - ( - m)} ) {}^{1 - m} \\ \\ = (a {}^{m + n} ) {}^{m - n} \times (a {}^{n + 1} ) {}^{n - 1} \times( a {}^{1 + m} ) {}^{1 - m} \\ \\ = a {}^{m {}^{2} - n {}^{2} } \times a {}^{n {}^{2} - 1{}^{2} } \times a {}^{1 {}^{2} - m {}^{2} } \\ \\ = a {}^{m {}^{2} - n {}^{2} + n {}^{2} - 1 {}^{2} + 1 {}^{2} - m {}^{2} } \\ \\ = a {}^{0} \\ \\ = 1 \: \: \: \: (proved)\\ \\ \\ formula \: - \: \: \: \: \frac{x {}^{m} }{x {}^{n} } = x {}^{m - n } \\ \\ x {}^{m} \times x {}^{n} = x {}^{m + n} \\ \\ (x {}^{m} ) {}^{n} = x {}^{mn} \\ \\ x {}^{0} = 1 \\ \\ (x + y)(x - y) = x {}^{2} - y {}^{2}$