Knowledge | alien547

a_{i} > 0 \Rightarrow \sum_{i = 1}^{n} a_{i} \geq n \cdot \sqrt[n]{\prod_{i = 1}^{n} a_{i}}

\sin^{2} θ + \cos^{2} θ = 1

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

a = \sqrt{b^{2} + c^{2} - 2 b c \cdot \cos α}

\frac{a}{\sin α} = \frac{b}{\sin β} = \frac{c}{\sin γ}

A_{n}^{m} = \frac{n!}{(n - m)!}

C_{n}^{m} = \frac{A_{n}^{m}}{m!}

| \begin{matrix} a_{11} & a_{12} & ⋯ & a_{1 n} \\ a_{21} & a_{22} & ⋯ & a_{2 n} \\ ⋮ & ⋮ & ⋮ \\ a_{n 1} & a_{n 2} & ⋯ & a_{n n} \end{matrix} | = \sum_{j_{1} j_{2} ⋯ j_{n}} {(- 1)}^{τ (j_{1} j_{2} ⋯ j_{n})} a_{1 j_{1}} a_{2 j_{2}} ⋯ a_{n j_{n}}

(a + b) (a - b) = a^{2} - b^{2}

{(a + b)}^{2} = a^{2} + 2 a b + b^{2}