প্রমাণ কর যে: $\dfrac{2^{2p+1}\cdot3^{2p+q}\cdot5^{p+q}\cdot6^{p}}{3^{p-2}\cdot6^{2p+2}\cdot10^{p}\cdot15^{q}}=\dfrac12$

Right Hand Side,
$=\dfrac{2^{2p+1}\cdot3^{2p+q}\cdot5^{p+q}\cdot6^{p}}{3^{p-2}\cdot6^{2p+2}\cdot10^{p}\cdot15^{q}}$

$=\dfrac{2^{2p+1}\cdot3^{2p+q}\cdot5^{p+q}\cdot\left(2\times3\right)^{p}}{3^{p-2}\cdot\left(2\times3\right)^{2p+2}\cdot\left(2\times5\right)^{p}\cdot\left(3\times5\right)^{q}}$

$=\dfrac{2^{2p+1}\cdot3^{2p+q}\cdot5^{p+q}\cdot2^{p}\cdot3^{p}}{3^{p-2}\cdot2^{2p+2}\cdot3^{2p+2}\cdot2^{p}\cdot5^{p}\cdot3^{q}\cdot5^{q}}$

$=\dfrac{2^{\left(2p+1\right)+p}\cdot3^{\left(2p+q\right)+p}\cdot5^{p+q}}{2^{\left(2p+2\right)+p}\cdot3^{\left(p-2\right)+\left(2p+2\right)+q}\cdot5^{p+q}}$

$=\dfrac{2^{2p+1+p}\cdot3^{2p+q+p}\cdot5^{p+q}}{2^{2p+2+p}\cdot3^{p-2+2p+2+q}\cdot5^{p+q}}$

$=\dfrac{2^{3p+1}\cdot3^{3p+q}\cdot5^{p+q}}{2^{3p+2}\cdot3^{3p+q}\cdot5^{p+q}}$

$=\dfrac{2^{3p+1}}{2^{3p+2}}\cdot\frac{3^{3p+q}}{3^{3p+q}}\cdot\frac{5^{p+q}}{5^{p+q}}$

$=\dfrac{2^{3p+1}}{2^{3p+2}}\cdot1\cdot1$

$=\dfrac{2^{3p+1}}{2^{3p+2}}$

$=2^{\left(3p+1\right)-\left(3p+2\right)}$

$=2^{3p+1-\left.3p-2\right)}$

$=2^{-1}$

$=\frac12$

$=$ Right Hand Side [Proved]