Question

Довести тотожність:
$\frac{2a - b}{ab} - \frac{1}{a + b}( \frac{a}{b} - \frac{b}{a}) = \frac{1}{b}$
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Answer 1

Доказательство:

$\frac{2a-b}{ab}-\frac{1}{a+b}(\frac{a}{b}-\frac{b}{a})=\frac{1}{b}\\\\\frac{2a-b}{ab}-\frac{1}{a+b}*\frac{a^2-b^2}{ab}=\frac{1}{b}\\\\\frac{2a-b}{ab}-\frac{1}{a+b}*\frac{(a-b)(a+b)}{ab}=\frac{1}{b}\\\\\frac{2a-b}{ab}-\frac{a-b}{ab}=\frac{1}{b}\\\\\frac{2a-b-a+b}{ab}=\frac{1}{b}\\\\\frac{a}{ab}=\frac{1}{b}\\\\\frac{1}{b}=\frac{1}{b}$

Тождество доказано

Answer 2

$\displaystyle\bf\\\frac{2a-b}{ab} -\frac{1}{a+b} \Big(\frac{a}{b} -\frac{b}{a} \Big)=\frac{2a-b}{ab} -\frac{1}{a+b}\cdot\frac{a^{2}-b^{2} }{ab} =\\\\\\=\frac{2a-b}{ab} -\frac{1}{a+b}\cdot\frac{(a-b)(a+b) }{ab} =\frac{2a-b}{ab} -\frac{a-b}{ab} =\\\\\\=\frac{2a-b-a+b}{ab} =\frac{a}{ab} =\frac{1}{b} \\\\\\\frac{1}{b} =\frac{1}{b}$

Что и требовалось доказать .