测试：论坛中直接编辑数学公式

wwwlib

$a_n*b^m$

trisinker

$3^2>2^2\DeltaABC~~\DeltaA^'B^'C^'$

没——问题

$\frac{1}{\sqrt 2} = \prod_{k=0}^\infty (1-\frac{1}{(4k+2)^2}) = (1-\frac{1}{4}) (1-\frac{1}{36}) (1-\frac{1}{100}) \cdots$ $\sqrt{2} = \prod_{k=0}^\infty \frac{(4k+2)^2}{(4k+1)(4k+3)} = (\frac{2 \cdot 2}{1 \cdot 3}) (\frac{6 \cdot 6}{5 \cdot 7}) (\frac{10 \cdot 10}{9 \cdot 11}) (\frac{14 \cdot 14}{13 \cdot 15}) \cdots$ $\sqrt{2} = \prod_{k=0}^\infty (1+\frac{1}{4k+1}) (1-\frac{1}{4k+3}) = (1+\frac{1}{1}) (1-\frac{1}{3}) (1+\frac{1}{5}) (1-\frac{1}{7}) \cdots.$ $\frac{1}{\sqrt{2}} = \sum_{k=0}^\infty \frac{(-1)^k (\frac{\pi}{4})^{2k}}{(2k)!}.$ $\sqrt{2} = \sum_{k=0}^\infty (-1)^{k+1} \frac{(2k-3)!!}{(2k)!!} = 1 + \frac{1}{2} - \frac{1}{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} - \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots.$ $\sqrt{2} = \sum_{k=0}^\infty \frac{(2k+1)!}{(k!)^2 2^{3k+1}} = \frac{1}{2} +\frac{3}{8} + \frac{15}{64} + \frac{35}{256} + \frac{315}{4096} + \frac{693}{16384} + \cdots.$ $\sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{\ddots}}}}. $

wayne

$(x-2)^2(y-2x+2)^2(y+2x-10)^2(x-4)^2(x-5)^2(y-2x+8)^2(y+2x-16)^2(x-7)^2(x-8)^2(y-2x+14)^2$ $\cdot(y+2x-22)^2(x-10)^2(x-13)^2((y-2)^2+|x-13|+|x-15|-2)^2((y-3)^2+|x-13|+|x-14|-1)^2$ $\cdot((y-4)^2+|x-13|+|x-15|-2)^2(x-16)^2(y+2x-36)^2(y-2x+32)^2(x-18)^2(y-2x+36)^2(y+2x-44)^2$ $\cdot((y-3)^2+|x-\frac{39}{2}|+|x-\frac{41}{2}|-1)^2(x-23)^2((y-4)^2+|x-22|+|x-24|-2)^2(x-25)^2(x-27)^2$ $\cdot((y-3)^2+|x-25|+|x-27|-2)^2(y-2x+58)^2(y+2x-66)^2((y-3)^2+|x-\frac{61}{2}|+|x-\frac{63}{2}|-1)^2$ $\cdot((y-\frac{1}{2}x+\frac{27}{2})^2+|x-33|+|x-35|-2)^2((y+\frac{1}{2}x-\frac{39}{2})^2+|x-33|+|x-35|-2)^2((y-\frac{1}{2}x+16)^2+|x-38|+|x-40|-2)^2$ $cdot((y+\frac{1}{2}x-22)^2+|x-38|+|x-40|-2)^2(x-41)^2(y+x-45)^2(x-43)^2+(y^2-6y+8+\sqrt{y^4-12y^3+52y^2-96y+64})^2=0$

只是呼吸

[tex]1^4-0^4=4\times1^3-6\times1^2+4\times1-1[tex] [tex]2^4-1^4=4\times2^3-6\times2^2+4\times2-1[tex]

只是呼吸

$1^4-0^4=4\times1^3-6\times1^2+4\times1-1$ $2^4-1^4=4\times2^3-6\times2^2+4\times2-1$

fzllz

test $\because$

clhu

还有一些符号显示不太正常，不知怎么回事。

xym_xuyimin

${\rm{C}}_n^k + {\rm{C}}_n^{k - 1} = {\rm{C}}_{n + 1}^k$

xym_xuyimin

$C_n^k + C_n^{k-1} =C_{n + 1}^k$