回复 9# 青青子衿
不截图了，太占位置，直接贴识别到的代码：
\iint_{D : x^{2}+y^{2} \leq 1 \atop D : x^{2} \leq y} d x d y=\int_{-\sqrt{\varphi}}^{\sqrt{\varphi}} \int_{x^{2}}^{\sqrt{1-x^{2}}} d y d x=\frac{\sqrt{\sqrt{5}-2}}{3}+\arcsin \sqrt{\frac{\sqrt{5}-1}{2}}
\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=\frac{1}{\sqrt{x^{2}+y^{2}}}
\left\{\begin{aligned} x^{3}+y &=10 \\+y^{3}+z &=10 \\ x &+z^{3}=10 \end{aligned}\right.
\left\{\begin{aligned}(-1+2 i)^{3}+(-1+2 i) &=10 \\+(-1+2 i)^{3}+(-1+2 i) &=10 \\(-1+2 i) &+(-1+2 i)^{3}=10 \end{aligned}\right.
\sum_{k=0}^{n}(-4)^{k} \left( \begin{array}{c}{n+k} \\ {2 k}\end{array}\right)效果：
$$\iint_{D : x^{2}+y^{2} \leq 1 \atop D : x^{2} \leq y} d x d y=\int_{-\sqrt{\varphi}}^{\sqrt{\varphi}} \int_{x^{2}}^{\sqrt{1-x^{2}}} d y d x=\frac{\sqrt{\sqrt{5}-2}}{3}+\arcsin \sqrt{\frac{\sqrt{5}-1}{2}}$$重积分下限变了样，d 也没自动直立了
$$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=\frac{1}{\sqrt{x^{2}+y^{2}}}$$正确
$$\left\{\begin{aligned} x^{3}+y &=10 \\+y^{3}+z &=10 \\ x &+z^{3}=10 \end{aligned}\right.$$
$$\left\{\begin{aligned}(-1+2 i)^{3}+(-1+2 i) &=10 \\+(-1+2 i)^{3}+(-1+2 i) &=10 \\(-1+2 i) &+(-1+2 i)^{3}=10 \end{aligned}\right.$$有空位的方程组的对齐看来不太行
$$\sum_{k=0}^{n}(-4)^{k} \left( \begin{array}{c}{n+k} \\ {2 k}\end{array}\right)$$并没有采用 \binom （咦？你咋改了） | 
kuing
发表于 2019-5-18 01:26
青青子衿
发表于 2019-5-18 17:22
abigtreenj
发表于 2019-10-23 21:23
不框住后面那个 =1725/1193 才能识别，而且也不会用 cfrac，只用了 frac：x=1+\frac{1}{2+\frac{1}{4+\frac{1}{8+\frac{1}{16}}}}
