\begin{gathered} k_{12}=-2.8881,k_{13}=2.888 \ k_{22}=0.4689,k_{23}=0.3099 \ k_{17}=-0.6953,k_{27}=0.1953 \end{gathered}已知g=9.81，m1,m2,m3,l1,l2均大于0，求m1,m2,m3,l1,l2怎么用matlab表达

根据题目中给出的公式和数值，可以列出以下方程组：

\begin{cases}
-2.8881 = \dfrac{3g(-2m_1 - 4(m_2+m_3))}{-2(4m_1+3(m_2+4m_3))l_1} \\
2.888 = \dfrac{-9gm_3}{-2(4m_1+3(m_2+4m_3))l_1} \\
0.4689 = \dfrac{2gm_2(m_1+2(m_2+m_3))l_1^2l_2}{4m_2^2l_1^2l_2^2-\dfrac{16}{9}m_2(m_1+3(m_2+m_3))l_1^2l_2^2} \\
0.3099 = \dfrac{-4gm_2(m_1+3(m_2+m_3))l_1^2l_2}{4m_2^2l_1^2l_2^2-\dfrac{16}{9}m_2(m_1+3(m_2+m_3))l_1^2l_2^2} \\
-0.6953 = \dfrac{3(-2m_1-m_1-4m_3)}{-2(4m_1+3(2m_2+4m_3))l_1} \\
0.1953 = \dfrac{2m_2(m_1+2(m_2+m_3))l_1^2l_2-\dfrac{4}{3}m_2(m_1+3(m_2+m_3))l_1^2l_2}{4m_2^2l_1^2l_2^2-\dfrac{16}{9}m_2(m_1+3(m_2+m_3))l_1^2l_2^2}
\end{cases}

将方程组中的常数和已知量代入Matlab中，可以用solve函数求解未知量：

g = 9.81;
k12 = -2.8881;
k13 = 2.888;
k22 = 0.4689;
k23 = 0.3099;
k17 = -0.6953;
k27 = 0.1953;

syms m1 m2 m3 l1 l2
[vars, vals] = solve([k12 == 3*g*(-2*m1-4*(m2+m3))/(-2*(4*m1+3*(m2+4*m3))*l1), ...
                      k13 == -9*g*m3/(-2*(4*m1+3*(m2+4*m3))*l1), ...
                      k22 == 2*g*m2*(m1+2*(m2+m3))*l1^2*l2/(4*m2^2*l1^2*l2^2-16/9*m2*(m1+3*(m2+m3))*l1^2*l2^2), ...
                      k23 == -4*g*m2*(m1+3*(m2+m3))*l1^2*l2/(4*m2^2*l1^2*l2^2-16/9*m2*(m1+3*(m2+m3))*l1^2*l2^2), ...
                      k17 == 3*(-2*m1-m1-4*m3)/(-2*(4*m1+3*(2*m2+4*m3))*l1), ...
                      k27 == (2*m2*(m1+2*(m2+m3))*l1^2*l2-4/3*m2*(m1+3*(m2+m3))*l1^2*l2)/(4*m2^2*l1^2*l2^2-16/9*m2*(m1+3*(m2+m3))*l1^2*l2^2)], ...
                     [m1, m2, m3, l1, l2]);
m1 = double(vals.m1);
m2 = double(vals.m2);
m3 = double(vals.m3);
l1 = double(vals.l1);
l2 = double(vals.l2);

执行完以上代码后，m1、m2、m3、l1、l2就分别是方程组的解。