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Question Number 193213 by simplifiedmaths965 last updated on 07/Jun/23
((a^m /a^(−n) ))^(m−n)
$$\left(\frac{{a}^{{m}} }{{a}^{−{n}} }\right)^{{m}−{n}} \\ $$
Answered by aba last updated on 07/Jun/23
((a^m /a^(−n) ))^(m−n) =(a^(m+n) )^(m−n) =a^(m^2 −n^2  )
$$\left(\frac{\mathrm{a}^{\mathrm{m}} }{\mathrm{a}^{−\mathrm{n}} }\right)^{\mathrm{m}−\mathrm{n}} =\left(\mathrm{a}^{\mathrm{m}+\mathrm{n}} \right)^{\mathrm{m}−\mathrm{n}} =\mathrm{a}^{\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} \:} \\ $$
Answered by Humble last updated on 07/Jun/23
a^((m−(−n))m−n)   a^((m+n)(m−n))     a^(m(m−n)+n(m−n))   a^(m^2 −nm+nm−n^2 )   a^(m^2 −n^2 )
$${a}^{\left({m}−\left(−{n}\right)\right){m}−{n}} \\ $$$${a}^{\left({m}+{n}\right)\left({m}−{n}\right)} \:\: \\ $$$${a}^{{m}\left({m}−{n}\right)+{n}\left({m}−{n}\right)} \\ $$$${a}^{{m}^{\mathrm{2}} \cancel{−{nm}+{nm}}−{n}^{\mathrm{2}} } \\ $$$${a}^{{m}^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$
Answered by MATHEMATICSAM last updated on 07/Jun/23
((a^m /a^(− n) ))^(m − n)   = (a^(m + n) )^(m − n)   = a^((m + n)(m − n))   = a^(m^2  − n^2 )
$$\left(\frac{{a}^{{m}} }{{a}^{−\:{n}} }\right)^{{m}\:−\:{n}} \\ $$$$=\:\left({a}^{{m}\:+\:{n}} \right)^{{m}\:−\:{n}} \\ $$$$=\:{a}^{\left({m}\:+\:{n}\right)\left({m}\:−\:{n}\right)} \\ $$$$=\:{a}^{{m}^{\mathrm{2}} \:−\:{n}^{\mathrm{2}} } \\ $$$$\: \\ $$

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