Question

Упрости выражение номер 13 14 15

Answer 1

$13)\; \; \frac{1}{1-a^2} \cdot \frac{a^2-1}{a+2} - \frac{2}{a^2+2a}= \frac{-(1-a^2)}{(1-a^2)(a+2)}- \frac{2}{a(a+2)}=\\\\= \frac{-a-2}{a(a+2)}=- \frac{a+2}{a(a+2)} =- \frac{1}{a}$

$14)\; \; \Big (\frac{n^2-an}{n^2-a^2}+\frac{a}{a+n}\Big ):\frac{a+n}{2} =\Big ( \frac{n(n-a)}{(n-a)(n+a)}+\frac{a}{a+n} \Big )\cdot \frac{2}{a+n} =\\\\=\Big ( \frac{n}{a+n} +\frac{a}{a+n}\Big )\cdot \frac{2}{a+n} = \frac{n+a}{a+n} \cdot \frac{2}{a+n}=1\cdot \frac{2}{a+n} =\frac{2}{a+n}$

$15)\; \; \Big (\frac{m}{m-n} -\frac{mn+n^2}{n^2-m^2}\Big )\cdot \frac{m-n}{(m+n)^2} =\Big ( \frac{m}{m-n} - \frac{n(m+n)}{(n-m)(n+m)}\Big )\cdot \frac{m-n}{(m+n)^2}=\\\\=\Big ( \frac{m}{m-n} -\frac{n}{n-m} \Big )\cdot \frac{m-n}{(m+n)^2} = \frac{m+n}{m-n} \cdot \frac{m-n}{(m+n)^2} = \frac{1}{m+n}$