Question

latex用aligned对齐后，给部分公式编号

\begin{eqnarray}\begin{aligned} {{\left( {{x}_{1}}-{{x}_{2}} \right)}^{T}}&\left[ {{f}^{R}}\left( t,{{x}_{1}},{{y}_{1}},{{x}_{1\tau }},{{y}_{1\tau }} \right)-{{f}^{R}}\left( t,{{x}_{2}},{{y}_{2}},{{x}_{2\tau }},{{y}_{2\tau }} \right) \right] \nonumber \\ &\le {{\left( {{x}_{1}}-{{x}_{2}} \right)}^{\text{T}}}L_{1}^{R}\left( {{x}_{1}}-{{x}_{2}} \right)+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{\text{T}}}L_{2}^{R}\left( {{y}_{1}}-{{y}_{2}} \right) \nonumber \\ &+{{\left( {{x}_{1\tau }}-{{x}_{2\tau }} \right)}^{T}}\Delta _{1}^{R}\left( {{x}_{1\tau }}-{{x}_{2\tau }} \right)+{{\left( {{y}_{1\tau }}-{{y}_{2\tau }} \right)}^{\text{T}}}\Delta _{2}^{R}\left( {{y}_{1\tau }}-{{y}_{2\tau }} \right), \\ {{\left( {{y}_{1}}-{{y}_{2}} \right)}^{T}}&\left[ {{f}^{I}}\left( t,{{x}_{1}},{{y}_{1}},{{x}_{1\tau }},{{y}_{1\tau }} \right)-{{f}^{I}}\left( t,{{x}_{2}},{{y}_{2}},{{x}_{2\tau }},{{y}_{2\tau }} \right) \right] \nonumber \\ &\le {{\left( {{x}_{1}}-{{x}_{2}} \right)}^{T}}L_{1}^{I}\left( {{x}_{1}}-{{x}_{2}} \right)+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{T}}L_{2}^{I}\left( {{y}_{1}}-{{y}_{2}} \right) \nonumber \\ &+ {{\left( {{x}_{1\tau }}-{{x}_{2\tau }} \right)}^{\text{T}}}\Delta _{1}^{I}\left( {{x}_{1\tau }}-{{x}_{2\tau }} \right)+{{\left( {{y}_{1\tau }}-{{y}_{2\tau }} \right)}^{\text{T}}}\Delta _{2}^{I}\left( {{y}_{1\tau }}-{{y}_{2\tau }} \right),\end{aligned} \end{eqnarray}
用其他对齐方式也行啊，能排出我的效果就可以呀，大神呀

Answer 1

直接这样就行

\begin{align}
{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{T}}&\left[ {{f}^{R}}\left( t,{{x}_{1}},{{y}_{1}},{{x}_{1\tau }},{{y}_{1\tau }} \right)-{{f}^{R}}\left( t,{{x}_{2}},{{y}_{2}},{{x}_{2\tau }},{{y}_{2\tau }} \right) \right] \nonumber \\
&\le {{\left( {{x}_{1}}-{{x}_{2}} \right)}^{\text{T}}}L_{1}^{R}\left( {{x}_{1}}-{{x}_{2}} \right)+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{\text{T}}}L_{2}^{R}\left( {{y}_{1}}-{{y}_{2}} \right) \nonumber \\
&+{{\left( {{x}_{1\tau }}-{{x}_{2\tau }} \right)}^{T}}\Delta _{1}^{R}\left( {{x}_{1\tau }}-{{x}_{2\tau }} \right)+{{\left( {{y}_{1\tau }}-{{y}_{2\tau }} \right)}^{\text{T}}}\Delta _{2}^{R}\left( {{y}_{1\tau }}-{{y}_{2\tau }} \right),\\
{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{T}}&\left[ {{f}^{I}}\left( t,{{x}_{1}},{{y}_{1}},{{x}_{1\tau }},{{y}_{1\tau }} \right)-{{f}^{I}}\left( t,{{x}_{2}},{{y}_{2}},{{x}_{2\tau }},{{y}_{2\tau }} \right) \right] \nonumber \\
&\le {{\left( {{x}_{1}}-{{x}_{2}} \right)}^{T}}L_{1}^{I}\left( {{x}_{1}}-{{x}_{2}} \right)+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{T}}L_{2}^{I}\left( {{y}_{1}}-{{y}_{2}} \right) \nonumber \\
&+ {{\left( {{x}_{1\tau }}-{{x}_{2\tau }} \right)}^{\text{T}}}\Delta _{1}^{I}\left( {{x}_{1\tau }}-{{x}_{2\tau }} \right)+{{\left( {{y}_{1\tau }}-{{y}_{2\tau }} \right)}^{\text{T}}}\Delta _{2}^{I}\left( {{y}_{1\tau }}-{{y}_{2\tau }} \right),
\end{align}

Answer 2

用gather环境加align应该可以自动编号，自定义编号用tag