यदि $\dfrac{x}{y} = \dfrac{z}{w}$ है, तो $\dfrac{x^{m} + y^{m} + z^{m} + w^{m}}{x^{-m} + y^{-m} + z^{-m} + w^{-m}} ?$
$\dfrac{x}{y}$
1
$(xyzw)^{m/2}$
$(xyzw)^{m}$
Solution
$\dfrac{x}{y} = \dfrac{z}{w} = k$ (मान लें)
$x = ky$ & $z = kw$
$\Rightarrow~\dfrac{x^{m} + y^{m} + z^{m} + w^{m}}{x^{-m} + y^{-m} + z^{-m} + w^{-m}}$
$= \dfrac{k^{m} y^{m} + y^{m} + k^{m} w^{m} + w^{m}}{k^{-m} y^{-m} + y^{-m} + k^{-m} w^{-m} + w^{-m}}$
$= \dfrac{y^{m} (k^{m} + 1) + w^{m} (k^{m} + 1)}{y^{-m} (k^{-m} + 1) + w^{-m} (k^{-m} + 1)}$
$= \dfrac{(k^{m} + 1) (y^{m} + w^{m})}{(k^{-m} + 1) (y^{-m} + w^{-m})}$
$= \dfrac{(k^{m} + 1) (y^{m} + w^{m})}{\left(\dfrac{1}{k^{m}} + 1\right) \left(\dfrac{1}{y^{m} + \dfrac{1}{w^3}}\right)}$
$= \dfrac{(k^{m} + 1) (y^{m} + w^{m})}{\dfrac{(k^{m} + 1)}{k^{m}} . \dfrac{(y^{m} + w^{m})}{y^{m}.w^{m}}}$
$= k^{m} y^{m} w^{m} = (kyw)^{w} = (k^2 y^2w^2)^{m/2}$
$= (ky. Y. w. kw)^{m/2} = (xyaw)^{m/2}$