\begin{aligned} &k_{12} ={\frac{3g(-2m_{1}-4(m_{2}+m_{3}))}{-2\left(4m_{1}+3(m_{2}+4m_{3})\right)l_{1}}},k_{13}={\frac{-9g m_{3}}{-2\left(4m_{1}+3\left(m_{2}+4m_{3}\right)\right)l_{1}}}. \ &k_{17} =\frac{3\left(-2m_1-m_1-4m_3\right)}{-2\left(4m_1+3\left(2m_2+4m_3\right)\right)l_1}, \ &k_{22} =\frac{\left(2g m_{2}(m_{1}+2(m_{2}+m_{3}))l_{1}^{2}l_{2}\right)}{4m_{2}^{2}l_{1}^{2}l_{2}^{2}-\frac{16}{9}m_{2}(m_{1}+3(m_{2}+m_{3}))l_{1}^{2}l_{2}^{2}}. \ &k_{23} =\frac{-\left(4g m_{2}(m_{1}+3(m_{2}+m_{3}))l_{1}^{2}l_{2}\right)}{4m_{2}^{2}l_{1}^{2}l_{2}^{2}-\frac{16}{9}m_{2}(m_{1}+3(m_{2}+m_{3}))l_{1}^{2}l_{2}^{2}}, \ &k_{27} ={\frac{2m_{2}(m_{1}+2(m_{2}+m_{3}))l_{1}^{2}l_{2}-{\frac{4}{3}}m_{2}(m_{1}+3(m_{2}+m_{3})l_{1}^{2}l_{2}}{4m_{2}^{2}l_{1}^{2}l_{2}^{2}-{\frac{16}{9}}m_{2}(m_{1}+3(m_{2}+m_{3}))l_{1}^{2}l_{2}^{2}}} \end{aligned}\begin{gathered} k_{12}=-2.8881,k_{13}=2.888 \ k_{22}=0.4689,k_{23}=0.3099 \ k_{17}=-0.6953,k_{27}=0.1958 \end{gathered}已知g=9.81,m1=0.5,m2=0.5,m3=0.25，求l1，l2

根据题目给出的方程组和已知条件，我们可以列出以下方程组：

$$
\begin{cases}
k_{12} = \dfrac{3g(-2m_{1}-4(m_{2}+m_{3}))}{-2\left(4m_{1}+3(m_{2}+4m_{3})\right)l_{1}} \\
k_{13}=\dfrac{-9g m_{3}}{-2\left(4m_{1}+3\left(m_{2}+4m_{3}\right)\right)l_{1}} \\
k_{17} = \dfrac{3\left(-2m_1-m_1-4m_3\right)}{-2\left(4m_1+3\left(2m_2+4m_3\right)\right)l_1} \\
k_{22} = \dfrac{\left(2g m_{2}(m_{1}+2(m_{2}+m_{3}))l_{1}^{2}l_{2}\right)}{4m_{2}^{2}l_{1}^{2}l_{2}^{2}-\frac{16}{9}m_{2}(m_{1}+3(m_{2}+m_{3}))l_{1}^{2}l_{2}^{2}} \\
k_{23} = \dfrac{-\left(4g m_{2}(m_{1}+3(m_{2}+m_{3}))l_{1}^{2}l_{2}\right)}{4m_{2}^{2}l_{1}^{2}l_{2}^{2}-\frac{16}{9}m_{2}(m_{1}+3(m_{2}+m_{3}))l_{1}^{2}l_{2}^{2}} \\
k_{27} = \dfrac{2m_{2}(m_{1}+2(m_{2}+m_{3}))l_{1}^{2}l_{2}-{\frac{4}{3}}m_{2}(m_{1}+3(m_{2}+m_{3})l_{1}^{2}l_{2}}{4m_{2}^{2}l_{1}^{2}l_{2}^{2}-{\frac{16}{9}}m_{2}(m_{1}+3(m_{2}+m_{3}))l_{1}^{2}l_{2}^{2}}
\end{cases}
$$

将已知数值带入，得到：

$$
\begin{cases}
k_{12} = -2.8881 \\
k_{13} = 2.888 \\
k_{17} = -0.6953 \\
k_{22} = 0.4689 \\
k_{23} = 0.3099 \\
k_{27} = 0.1958 \\
g = 9.81 \\
m_1 = 0.5 \\
m_2 = 0.5 \\
m_3 = 0.25 \\
\end{cases}
$$

将方程组化简，得到：

$$
\begin{cases}
-3.7037l_1 - 0.6667m_2l_2 = 0 \\
0.8889l_1 - 2.25m_2l_2 = 0 \\
-1.3889l_1 + 0.4444m_2l_2 = 0
\end{cases}
$$

解这个方程组，得到：

$$
\begin{cases}
l_1 = 0 \\
l_2 = 0
\end{cases}
$$

但这个结果明显不符合物理实际，因此我们考虑方程组无解的情况。可以发现，第1个和第3个方程左边都是$l_1$的线性组合，而第2个方程左边不含$l_1$，因此这个方程组无解的充要条件是：

$$
\begin{bmatrix}
-3.7037 & -0.6667m_2 \\
0.8889 & -2.25m_2 \\
-1.3889 & 0.4444m_2
\end{bmatrix}
\begin{bmatrix}
l_1 \\
l_2
\end{bmatrix}
$$

的秩小于2。计算可得：

$$
\begin{vmatrix}
-3.7037 & -0.6667m_2 \\
0.8889 & -2.25m_2 \\
-1.3889 & 0.4444m_2
\end{vmatrix} = -\frac{5}{18}m_2^2
$$

因此，当 $m_2 \neq 0$ 时，方程组无解。当 $m_2 = 0$ 时，根据方程组可以得到：

$$
\begin{cases}
l_1 = 0 \\
l_2 = \text{任意实数}
\end{cases}
$$

综上所述，当 $m_2 = 0$ 时，$l_1 = 0$，$l_2$ 为任意实数；当 $m_2 \neq 0$ 时，方程组无解。