求解 {l}{8=a^2-b^2}{5/9n^2+4/b^2=1}

$$\left\{ \begin{array} { l } { 8 = a ^ { 2 } - b ^ { 2 } } \\ { \frac { 5 } { 9 n ^ { 2 } } + \frac { 4 } { b ^ { 2 } } = 1 } \end{array} \right.$$

求解 a, b, n 的值

$$a=2i\left(-9+\frac{5}{n^{2}}\right)^{-\frac{1}{2}}\sqrt{27-\frac{10}{n^{2}}}\text{, }b=6i\left(-9+\frac{5}{n^{2}}\right)^{-\frac{1}{2}}\text{, }n\in \mathrm{C}\setminus 0,-\frac{\sqrt{5}}{3},\frac{\sqrt{5}}{3}$$

$$a=2i\left(-9+\frac{5}{n^{2}}\right)^{-\frac{1}{2}}\sqrt{27-\frac{10}{n^{2}}}\text{, }b=-6i\left(-9+\frac{5}{n^{2}}\right)^{-\frac{1}{2}}\text{, }n\in \mathrm{C}\setminus 0,-\frac{\sqrt{5}}{3},\frac{\sqrt{5}}{3}$$

$$a=-2i\left(-9+\frac{5}{n^{2}}\right)^{-\frac{1}{2}}\sqrt{27-\frac{10}{n^{2}}}\text{, }b=6i\left(-9+\frac{5}{n^{2}}\right)^{-\frac{1}{2}}\text{, }n\in \mathrm{C}\setminus 0,-\frac{\sqrt{5}}{3},\frac{\sqrt{5}}{3}$$

$$a=-2i\left(-9+\frac{5}{n^{2}}\right)^{-\frac{1}{2}}\sqrt{27-\frac{10}{n^{2}}}\text{, }b=-6i\left(-9+\frac{5}{n^{2}}\right)^{-\frac{1}{2}}\text{, }n\in \mathrm{C}\setminus 0,-\frac{\sqrt{5}}{3},\frac{\sqrt{5}}{3}$$