Question

распишите подробно решение.​

Answer 1

a)

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

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

б)

$\frac{m-n}{m+n} + \frac{m+n}{m-n} = \frac{(m-n)(m-n) + (m+n)(m+n)}{(m+n)(m-n)} = \frac{m^{2} - 2mn + n^{2} + m^{2} +2mn+n^{2} }{m^{2}-n^{2} } = \frac{2m^{2} + 2n^{2} }{m^{2}-n^{2} } = \frac{2(m^{2}+n^{2})}{m^{2}-n^{2}}$

$\frac{m-n}{m+n} + \frac{m+n}{m-n} = \frac{(m-n)(m-n) + (m+n)(m+n)}{(m+n)(m-n)} = \frac{m^{2} - 2mn + n^{2} - m^{2} -2mn-n^{2} }{m^{2}-n^{2} } = \frac{-4mn }{m^{2}-n^{2} }$