а)
\(\frac{a^2-b^2}{a^2+2ab+b^2} = \frac{(a-b)(a+b)}{(a+b)^2} = \frac{a-b}{a+b}\)При \(a=\frac{2}{3}, b=\frac{1}{3}\):
\(\frac{\frac{2}{3}-\frac{1}{3}}{\frac{2}{3}+\frac{1}{3}} = \frac{\frac{1}{3}}{\frac{3}{3}} = \frac{\frac{1}{3}}{1} = \frac{1}{3}\)б)\(\frac{a^2-2ab+b^2}{a^2-b^2} = \frac{(a-b)^2}{(a-b)(a+b)} = \frac{a-b}{a+b}\)При \(a=\frac{4}{7}, b=\frac{3}{7}\):
\(\frac{\frac{4}{7}-\frac{3}{7}}{\frac{4}{7}+\frac{3}{7}} = \frac{\frac{1}{7}}{\frac{7}{7}} = \frac{\frac{1}{7}}{1} = \frac{1}{7}\)Ответ: а) \(\frac{1}{3}\), б) \(\frac{1}{7}\).