Question

please solve it
$({(a + 1 \div b)}^{m} \times {(a - 1 \div b)}^{n} ) \div ( {(b + 1 \div a)}^{m} \times ( {b - 1 \div a)}^{n} = a \div b$
please solve it​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \dfrac{{\bigg(a + \dfrac{1}{b} \bigg) }^{m}{\bigg(a - \dfrac{1}{b} \bigg) }^{n}}{{\bigg(b + \dfrac{1}{a} \bigg) }^{m}{\bigg(b - \dfrac{1}{a} \bigg) }^{n}} = \dfrac{a}{b}$

can be rewritten as on taking LCM in each bracket,

$\rm \: \dfrac{{\bigg(\dfrac{ab + 1}{b} \bigg) }^{m}{\bigg(\dfrac{ab - 1}{b} \bigg) }^{n}}{{\bigg(\dfrac{ab + 1}{a} \bigg) }^{m}{\bigg(\dfrac{ab - 1}{a} \bigg) }^{n}} = \dfrac{a}{b}$

We know,

$\boxed{\tt{ {\bigg(\dfrac{x}{y} \bigg) }^{n} = \frac{ {x}^{n} }{ {y}^{n} } \: }}$

So, using this identity, we get

$\rm \: \dfrac{\dfrac{ {(ab + 1)}^{m} }{ {b}^{m} } \dfrac{ {(ab - 1)}^{n} }{ {b}^{n} } }{\dfrac{ {(ab + 1)}^{m} }{ {a}^{m} } \dfrac{ {(ab - 1)}^{n} }{ {a}^{n} } } = \dfrac{a}{b}$

So, after canceling the like terms, we get

$\rm \: \dfrac{ {a}^{m} \times {a}^{n} }{ {b}^{m} \times {b}^{n} } = \dfrac{a}{b}$

We know,

$\boxed{\tt{ {x}^{m} \times {x}^{n} = {x}^{m + n} }}$

So, using this, we get

$\rm \: \dfrac{ {a}^{m + n} }{ {b}^{m + n} } = \dfrac{a}{b}$

$\rm \: {\bigg(\dfrac{a}{b} \bigg) }^{m + n} = \dfrac{a}{b}$

$\rm \: {\bigg(\dfrac{a}{b} \bigg) }^{m + n} = {\bigg(\dfrac{a}{b} \bigg) }^{1}$

We know,

$\boxed{\tt{ {x}^{m} = {x}^{n} \: \: \rm\implies \:m = n \: }}$

So, using this, we get

$\rm \: m + n = 1$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\boxed{\tt{ {x}^{0} = 1 \: }}$

$\boxed{\tt{ {x}^{ - n} = \frac{1}{ {x}^{n} } \: }}$

$\boxed{\tt{ {\bigg(\dfrac{x}{y} \bigg) }^{ - n} = {\bigg(\dfrac{y}{x} \bigg) }^{n} \: }}$

$\boxed{\tt{ {x}^{m} \times {y}^{m} = {(xy)}^{m} \: }}$

$\boxed{\tt{ {x}^{m} \div {y}^{m} = {\bigg(\dfrac{x}{y} \bigg) }^{m} \: }}$

$\boxed{\tt{ {0}^{0} \: is \: not \: defined \: }}$

Answer 2

Step-by-step explanation:

$\huge\mathcal\colorbox{yellow}{{\color{cyan}{\boxed{\huge{\mathbb\purple{REFERR \:TO\: THE\:\: ATTACHMENT }}}}}}$

Answer: