Доказательство:

[tex]\frac{2a-b}{ab}-\frac{1}{a+b}(\frac{a}{b}-\frac{b}{a})=\frac{1}{b}\\\\\frac{2a-b}{ab}-\frac{1}{a+b}*\frac{a^2-b^2}{ab}=\frac{1}{b}\\\\\frac{2a-b}{ab}-\frac{1}{a+b}*\frac{(a-b)(a+b)}{ab}=\frac{1}{b}\\\\\frac{2a-b}{ab}-\frac{a-b}{ab}=\frac{1}{b}\\\\\frac{2a-b-a+b}{ab}=\frac{1}{b}\\\\\frac{a}{ab}=\frac{1}{b}\\\\\frac{1}{b}=\frac{1}{b}[/tex]

Тождество доказано

[tex]\displaystyle\bf\\\frac{2a-b}{ab} -\frac{1}{a+b} \Big(\frac{a}{b} -\frac{b}{a} \Big)=\frac{2a-b}{ab} -\frac{1}{a+b}\cdot\frac{a^{2}-b^{2} }{ab} =\\\\\\=\frac{2a-b}{ab} -\frac{1}{a+b}\cdot\frac{(a-b)(a+b) }{ab} =\frac{2a-b}{ab} -\frac{a-b}{ab} =\\\\\\=\frac{2a-b-a+b}{ab} =\frac{a}{ab} =\frac{1}{b} \\\\\\\frac{1}{b} =\frac{1}{b}[/tex]

Что и требовалось доказать .