Question

Доведіть тотожність : (5/a2+2ab+b2 + 1/a2-b2 - 6/a2-2ab+b2): 11a+b/a2-b2= -2b/a2-b2

Answer 1

$.......= (\frac{5}{(a+b)(a+b) } + \frac{1}{(a+b)(a-b)} - \frac{6}{(a-b)(a-b)})* \frac{(a+b)(a-b)}{11a+b} =$ $\frac{5(a-b)^{2}+(a+b)(a-b)-6(a+b)^{2}}{(a+b)^{2}(a-b)^{2}} * \frac{(a+b)(a-b)}{11a+b}=$ $\frac{5a^{2}-10ab+5b^{2}+a^{2} -b^{2}-6a^{2}-12ab-6b^{2}}{(a+b)(a-b)(11a+b)}= \frac{-2b^{2}-22ab }{(a^{2}-b^{2})(11a+b)}= \frac{-2b(b+11a)}{(a^2-b^2)(11a+b)}=$ $-\frac{2b}{a^2-b^2}$
Доказано.