求解 frac{k_{1}+1}{k_1^2+2}+frac{k_{2}+1}{k_2^2+2}=24/16

$$\frac { k _ { 1 } + 1 } { k _ { 1 } ^ { 2 } + 2 } + \frac { k _ { 2 } + 1 } { k _ { 2 } ^ { 2 } + 2 } = \frac { 24 } { 16 }$$

求解 k_1 的值

$$k_{1}=\frac{\sqrt{-11k_{2}^{4}+20k_{2}^{3}-32k_{2}^{2}+24k_{2}-12}-k_{2}^{2}-2}{-3k_{2}^{2}+2k_{2}-4}$$

$$k_{1}=-\frac{\sqrt{-11k_{2}^{4}+20k_{2}^{3}-32k_{2}^{2}+24k_{2}-12}+k_{2}^{2}+2}{-3k_{2}^{2}+2k_{2}-4}\text{, }-11k_{2}^{4}+20k_{2}^{3}-32k_{2}^{2}+24k_{2}-12\geq 0$$

求解 k_2 的值

$$k_{2}=\frac{\sqrt{-11k_{1}^{4}+20k_{1}^{3}-32k_{1}^{2}+24k_{1}-12}-k_{1}^{2}-2}{-3k_{1}^{2}+2k_{1}-4}$$

$$k_{2}=-\frac{\sqrt{-11k_{1}^{4}+20k_{1}^{3}-32k_{1}^{2}+24k_{1}-12}+k_{1}^{2}+2}{-3k_{1}^{2}+2k_{1}-4}\text{, }-11k_{1}^{4}+20k_{1}^{3}-32k_{1}^{2}+24k_{1}-12\geq 0$$