\[\boxed{\text{159\ (159).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Упростим\ левую\ часть\ \]
\[равенства:\]
\[\frac{2p - q}{\text{pq}} - \frac{1}{p + q} \cdot \frac{p^{2} - q^{2}}{\text{pq}} = \frac{1}{q}\]
\[\frac{2p - q}{\text{pq}} - \frac{p - q}{\text{pq}} = \frac{1}{q}\]
\[\frac{2p - q - p + q}{\text{pq}} = \frac{1}{q}\]
\[\frac{p}{\text{pq}} = \frac{1}{q}\]
\[\frac{1}{q} = \frac{1}{q}.\]
\[Что\ и\ требовалось\ доказать.\]
\[\textbf{б)}\ \frac{a + b^{\backslash a + b}}{2(a - b)} - \frac{a - b^{\backslash a - b}}{2(a + b)} =\]
\[= \frac{b^{\backslash a + b}}{a - b} - \frac{b^{2} - ab}{a^{2} - b^{2}}\]
\[Упростим\ обе\ части\ равенства:\]
\[= \frac{ab + b^{2} - b^{2} + ab}{a^{2} - b^{2}}\]
\[\frac{4ab}{2\left( a^{2} - b^{2} \right)} = \frac{2ab}{a^{2} - b^{2}}\]
\[\frac{2ab}{a^{2} - b^{2}} = \frac{2ab}{a^{2} - b^{2}}\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]