1.已知$AX=B$,其中$A=\left[\begin{array}{ll}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right], B=\left[\begin{array}{ccc}{2} & {5} & {-1} \\ {4} & {10} & {-2} \\ {7} & {9} & {3}\end{array}\right]$,求$X$.
法一.由于\[\left[\begin{array}{cc}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right]\left[\begin{array}{lll}{x_{11}} & {x_{12}} & {x_{13}} \\ {x_{21}} & {x_{22}} & {x_{23}}\end{array}\right]=\left[\begin{array}{ccc}{2} & {5} & {-1} \\ {4} & {10} & {-2} \\ {7} & {9} & {3}\end{array}\right],\]
则|\[\left[\begin{array}{ll}{1} & {2} \\ {3} & {5}\end{array}\right]\left[\begin{array}{lll}{x_{11}} & {x_{12}} & {x_{13}} \\ {x_{21}} & {x_{22}} & {x_{23}}\end{array}\right]=\left[\begin{array}{ccc}{2} & {5} & {-1} \\ {7} & {9} & {3}\end{array}\right],\]
于是\[\left[\begin{array}{ccc}{x_{11}} & {x_{12}} & {x_{13}} \\ {x_{21}} & {x_{22}} & {x_{23}}\end{array}\right]=\left[\begin{array}{cc}{1} & {2} \\ {3} & {5}\end{array}\right]^{-1}\left[\begin{array}{ccc}{2} & {5} & {-1} \\ {7} & {9} & {3}\end{array}\right]=\left[\begin{array}{ccc}{4} & {-7} & {11} \\ {-1} & {6} & {-6}\end{array}\right].\]
法二.注意到\[\left[\begin{array}{cc}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right]\left[\begin{array}{l}{x_{11}} \\ {x_{21}}\end{array}\right]=\left[\begin{array}{l}{2} \\ {4} \\ {7}\end{array}\right],\left[\begin{array}{ll}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right]\left[\begin{array}{l}{x_{12}} \\ {x_{22}}\end{array}\right]=\left[\begin{array}{c}{5} \\ {10} \\ {9}\end{array}\right],\left[\begin{array}{cc}{1} & {2} \\ {2} & {4} \\ {3} & {5}\end{array}\right]\left[\begin{array}{c}{x_{13}} \\ {x_{23}}\end{array}\right]=\left[\begin{array}{c}{-1} \\ {-2} \\ {3}\end{array}\right],\]
则\[\left[\begin{array}{l}{x_{11}} \\ {x_{21}}\end{array}\right]=\left[\begin{array}{l}{4} \\ {-1}\end{array}\right],\quad\left[\begin{array}{l}{x_{12}} \\ {x_{22}}\end{array}\right]=\left[\begin{array}{c}{-7} \\ {6}\end{array}\right],\quad\left[\begin{array}{l}{x_{13}} \\ {x_{23}}\end{array}\right]=\left[\begin{array}{c}{11} \\ {-6}\end{array}\right].\]因此\[X=\left[\begin{array}{ccc}{4} & {-7} & {11} \\ {-1} & {6} & {-6}\end{array}\right].\]
常用级数\[(1+x)^{1/x}=e-\frac{e x}{2}+\frac{11 e x^{2}}{24}-\frac{7 e x^{3}}{16}+\frac{2447 e x^{4}}{5760}+O\left(x^{5}\right),\quad x\to 0\]
\[\int_{0}^{1} \sqrt{\frac{x}{1-x}} d x=\frac{\pi}{2}.\]
原文：https://www.cnblogs.com/Eufisky/p/10987674.html