解決{l}{8=a^2-b^2}{5/9n^2+4/b^2=1} |Microsoft數學求解器

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

解 a、b、n (復數求解)

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

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

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

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

解 a、b、n

$$a=-2\sqrt{\frac{27n^{2}-10}{9n^{2}-5}}\text{, }b=-6\sqrt{-\frac{1}{5-9n^{2}}}|n|\text{, }n\in \left(-\infty,-\frac{\sqrt{5}}{3}\right)\cup \left(\frac{\sqrt{5}}{3},\infty\right)$$

$$a=-2\sqrt{\frac{27n^{2}-10}{9n^{2}-5}}\text{, }b=6\sqrt{-\frac{1}{5-9n^{2}}}|n|\text{, }n\in \left(-\infty,-\frac{\sqrt{5}}{3}\right)\cup \left(\frac{\sqrt{5}}{3},\infty\right)$$

$$a=2\sqrt{\frac{27n^{2}-10}{9n^{2}-5}}\text{, }b=-6\sqrt{-\frac{1}{5-9n^{2}}}|n|\text{, }n\in \left(-\infty,-\frac{\sqrt{5}}{3}\right)\cup \left(\frac{\sqrt{5}}{3},\infty\right)$$

$$a=2\sqrt{\frac{27n^{2}-10}{9n^{2}-5}}\text{, }b=6\sqrt{-\frac{1}{5-9n^{2}}}|n|\text{, }n\in \left(-\infty,-\frac{\sqrt{5}}{3}\right)\cup \left(\frac{\sqrt{5}}{3},\infty\right)$$