结果

结果

代码

结果

E=mc^2

Inline 行内的公式 E=mc^2 行内的公式，行内的E=mc^2公式。

c = \pm\sqrt{a^2 + b^2}

x > y

f(x) = x^2

\alpha = \sqrt{1-e^2}

(\sqrt{3x-1}+(1+x)^2)

\sin(\alpha)^{\theta}=\sum_{i=0}^{n}(x^i + \cos(f))

\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

f(x) = \int_{-\infty}^\infty\hat f(\xi)\,e^{2 \pi i \xi x}\,d\xi

\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }

\displaystyle \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)

a^2

a^{2+2}

a_2

x_2^3

10^{10^{8}}

a_{i,j}

_nP_k

c = \pm\sqrt{a^2 + b^2}

\frac{1}{2}=0.5

\dfrac{k}{k-1} = 0.5

\dbinom{n}{k} \binom{n}{k}

\oint_C x^3\, dx + 4y^2\, dy

\bigcap_1^n p \bigcup_1^k p

e^{i \pi} + 1 = 0

\left ( \frac{1}{2} \right )

x_{1,2}=\frac{-b\pm\sqrt{\color{Red}b^2-4ac}}{2a}

{\color{Blue}x^2}+{\color{YellowOrange}2x}-{\color{OliveGreen}1}

\textstyle \sum_{k=1}^N k^2

\dfrac{ \tfrac{1}{2}[1-(\tfrac{1}{2})^n] }{ 1-\tfrac{1}{2} } = s_n

0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+\cdots

\sum_{k=1}^N k^2

\textstyle \sum_{k=1}^N k^2

\prod_{i=1}^N x_i

\textstyle \prod_{i=1}^N x_i

\coprod_{i=1}^N x_i

\textstyle \coprod_{i=1}^N x_i

\int_{1}^{3}\frac{e^3/x}{x^2}\, dx

\int_C x^3\, dx + 4y^2\, dy

{}_1^2!\Omega_3^4

多行公式 Multi line

代码

结果

f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi

\displaystyle \left( \sum\_{k=1}^n a\_k b\_k \right)^2 \leq \left( \sum\_{k=1}^n a\_k^2 \right) \left( \sum\_{k=1}^n b\_k^2 \right)

\dfrac{ \tfrac{1}{2}[1-(\tfrac{1}{2})^n] } { 1-\tfrac{1}{2} } = s_n

\displaystyle \frac{1}{ \Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{ \frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} { 1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }

f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi

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