Question Number 172820 by Mikenice last updated on 01/Jul/22
Answered by CElcedricjunior last updated on 02/Jul/22
![(((x+(1/y))^m (x−(1/y))^n )/((y+(1/x))^m (y−(1/x))^n ))=(((x+(1/y))/(y+(1/x))))^m ×(((x−(1/y))/(y−(1/x))))^n =[(((xy+1)/y))×((x/(xy+1)))]^n ×[(((xy−1)/y))×((x/(xy−1)))]^m =(x^(n+m) /y^(n+m) )=((x/y))^(n+m) ...........Le ce^ le^ bre cedric junior...........](https://www.tinkutara.com/question/Q172865.png)
$$\frac{\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}\right)^{\boldsymbol{\mathrm{m}}} \left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}\right)^{\boldsymbol{\mathrm{n}}} }{\left(\boldsymbol{\mathrm{y}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\boldsymbol{\mathrm{m}}} \left(\boldsymbol{\mathrm{y}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\boldsymbol{\mathrm{n}}} }=\left(\frac{\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}}{\boldsymbol{\mathrm{y}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}}\right)^{\boldsymbol{\mathrm{m}}} ×\left(\frac{\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}}{\boldsymbol{\mathrm{y}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$$$=\left[\left(\frac{\boldsymbol{\mathrm{xy}}+\mathrm{1}}{\boldsymbol{\mathrm{y}}}\right)×\left(\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{xy}}+\mathrm{1}}\right)\right]^{\boldsymbol{\mathrm{n}}} ×\left[\left(\frac{\boldsymbol{\mathrm{xy}}−\mathrm{1}}{\boldsymbol{\mathrm{y}}}\right)×\left(\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{xy}}−\mathrm{1}}\right)\right]^{\boldsymbol{\mathrm{m}}} \\ $$$$=\frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}} }{\boldsymbol{\mathrm{y}}^{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}} }=\left(\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}\right)^{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}} \\ $$$$ \\ $$$$ \\ $$$$………..\mathscr{L}{e}\:{c}\acute {{e}l}\grave {{e}bre}\:{cedric}\:{junior}……….. \\ $$
Commented by peter frank last updated on 02/Jul/22
$$\mathrm{good} \\ $$