这个页面用于练习TeX数学公式排版。
T h i s i s a ¯ P a g e f o r ¨ T E X t e s t i n g ⏟ {\displaystyle {\boldsymbol {This}}\ {\overline {is\ a}}\ P^{age}\ fo{\ddot {r}}\ \underbrace {\mathrm {T_{E}X} \ {\mathfrak {testing}}} }
H 2 C O 3 = = = Δ H 2 O + C O 2 ↑ {\displaystyle \mathrm {H_{2}CO_{3}} \ {\overset {\Delta }{=\!=\!=}}\ \mathrm {H_{2}O} +\mathrm {CO_{2}} \uparrow }
这个算式对任意正整数 n {\displaystyle n} 都成立： 1 + 1 n 2 + 1 ( n + 1 ) 2 = n 2 ( n + 1 ) 2 + n 2 + ( n + 1 ) 2 n 2 ( n + 1 ) 2 = [ n ( n + 1 ) ] 2 + 2 n ( n + 1 ) + 1 [ n ( n + 1 ) ] 2 = ( n 2 + n + 1 ) 2 [ n ( n + 1 ) ] 2 = n 2 + n + 1 n ( n + 1 ) = 1 + 1 n ( n + 1 ) {\displaystyle {\begin{aligned}{\sqrt {1+{\frac {1}{n^{2}}}+{\frac {1}{(n+1)^{2}}}}}&={\sqrt {\frac {n^{2}(n+1)^{2}+n^{2}+(n+1)^{2}}{n^{2}(n+1)^{2}}}}\\&={\sqrt {\frac {[n(n+1)]^{2}+2n(n+1)+1}{[n(n+1)]^{2}}}}\\&={\sqrt {\frac {(n^{2}+n+1)^{2}}{[n(n+1)]^{2}}}}\\&={\frac {n^{2}+n+1}{n(n+1)}}\\&=1+{\frac {1}{n(n+1)}}\end{aligned}}}
义勇军进行曲
1 = G 2 4 {\displaystyle 1=\mathrm {G} \qquad {\tfrac {2}{4}}}
( 1. _ 3 _ _ 5 5 _ | 6 5 | 3. _ 1 _ _ 5 5 5 ⌢ 3 _ | 3 1 | 5 _ ˙ 5 _ ˙ 5 _ ˙ ⌢ 3 5 _ ˙ 5 _ ˙ 5 _ ˙ ⌢ 3 | 1 ) 0 _ 5 _ ˙ | 1. 1 _ | {\displaystyle {\begin{smallmatrix}({\underline {1.\;}}{\underline {\underline {3}}}\ {\underline {5\;5}}\;|\;6\quad 5\;|\;{\underline {3.\;}}{\underline {\underline {1}}}\ {\underline {\overset {\overset {3}{\frown }}{5\;5\;5}}}\;|\;3\quad 1\;|\;{\overset {\overset {3}{\frown }}{{\underset {{\dot {}}\;}{\underline {5\;}}}{\underset {{\dot {}}\;}{\underline {5\;}}}{\underset {\dot {}}{\underline {5}}}}}\;{\overset {\overset {3}{\frown }}{{\underset {{\dot {}}\;}{\underline {5\;}}}{\underset {{\dot {}}\;}{\underline {5\;}}}{\underset {\dot {}}{\underline {5}}}}}\;|\;1)\end{smallmatrix}}\;{\underline {0\;}}{\underset {\dot {}}{\underline {5}}}\;|\;1.\qquad {\underline {1}}\;|}
都成立：
![{\displaystyle {\begin{aligned}{\sqrt {1+{\frac {1}{n^{2}}}+{\frac {1}{(n+1)^{2}}}}}&={\sqrt {\frac {n^{2}(n+1)^{2}+n^{2}+(n+1)^{2}}{n^{2}(n+1)^{2}}}}\\&={\sqrt {\frac {[n(n+1)]^{2}+2n(n+1)+1}{[n(n+1)]^{2}}}}\\&={\sqrt {\frac {(n^{2}+n+1)^{2}}{[n(n+1)]^{2}}}}\\&={\frac {n^{2}+n+1}{n(n+1)}}\\&=1+{\frac {1}{n(n+1)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7058ba585825231e00d52f610ec7f2bff8ddebf)
