由已知条件，得到m=log7 11，n=log11 7，带入求解方程组，得\frac{m}{m+1} + \frac{n}{n+1} = \frac{log7 11}{log7 11 + 1} + \frac{log11 7}{log11 7 + 1} = \frac{log(7^{log7 11})}{log(7^{log7}) + log(7^{log11})} = \frac{log(7^{log7 log11})}{log(7) + log(11)} = \frac{log(11)}{log(7) + log(11)} = \frac{lg 2}{lg3 + lg4} = \frac{lg8}{lg9}
故得\frac{m}{m+1} + \frac{n}{n+1} = \frac{lg8}{lg9}
其中涉及到的换底公式公式推导：\frac{g_b^{m}}{g_a^{m}} = \frac{b^m}{\frac{b^{a}}{a^m}} = \frac{b^{a}}{a^{a}} \cdot a = \frac{a}{b}
所以最终结果为\frac{lg9}{lg8}
其中lg代表以10为底的对数。