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$${(a\pm b)}^2=a^2\pm 2ab+b^2{(a\pm b)}^3=a^3\pm 3a^{2}b+3a{b}^2\pm b^3{(a+b)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^k{b}^{n-k}{a}^2-b^2=(a-b)(a+b)a^3-b^3=(a-b)(a^2+ab+b^2)$$
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +a{b}^{n-2}+b^{n-1})a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)(a^2+n{b}^2)(c^2+n{d}^2)={(ac\pm nbd)}^2+n{(ad\mp bc)}^2$$
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)a{x}^2+bx+c=0⟺x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}{D}_2=b^2-4ac{\alpha }_2+{\beta }_2=-\frac{a}{b}{\alpha }_2{\beta }_2=\frac{c}{a}$$
$$y^3+ay+b=0⟺y=\frac{1}{3}(\sqrt[3]{-\frac{27}{2}b+\frac{3\sqrt{3}i}{2}\sqrt{-4a^3-27b^2}}-\frac{3a}{\sqrt[3]{-\frac{27}{2}b+\frac{3\sqrt{3}i}{2}\sqrt{-4a^3-27b^2}-}})(y\to x-\frac{b}{3})$$
$$D_3=-4a{c}^3+b^2{c}^2-4b^{2}d+18abcd-27a^2{d}^2{\alpha }_3+{\beta }_3+{\gamma }_3=-\frac{b}{a}{\alpha }_3{\beta }_3+{\beta }_3{\gamma }_3+{\gamma }_3{\alpha }_3=\frac{c}{a}{\alpha }_3{\beta }_3{\gamma }_3=-\frac{d}{a}\frac{a+b}{2}\ge \sqrt{ab}{a}^2+b^2+c^2\ge ab+bc+ca$$
$$\sum _{i=1}^n{a}_i\ge n\sqrt[n]{\prod _{i=1}^n{a}_i}\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}a\sqrt{x}+b\sqrt{y}\le \sqrt{(a^2+b^2)(x+y)}\log x\le x-1{(1+x)}^n\ge 1+nx(a_1^2+a_2^2)(b_1^2+b_2^2)\ge {(a_1{b}_1+a_2{b}_2)}^2$$
$$(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)\ge {(a_1{b}_1+a_2{b}_2+a_3{b}_3)}^2\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)\ge {\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2\frac{\pi }{2}x\le \sin x\le x(0\ge x\ge \frac{\pi }{2}){\sin }^2\theta +{\cos }^2\theta =11+{\tan }^2\theta =\frac{1}{{\cos }^2\theta }$$
$$\sin (a\pm b)=\sin a\cos b\pm \cos a\sin b\cos (a\pm b)=\cos a\cos b\mp \sin a\sin b\tan (a\pm b)=\frac{\tan a\pm \tan b}{1\mp \tan a\tan b}\sin 2\theta =2\sin \theta \cos \theta \cos 2\theta =2{\cos }^2\theta -1$$
$$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\sin 3\theta =-4{\sin }^3\theta +3\sin \theta \cos 3\theta =4{\cos }^3\theta -3\cos \theta \tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }{\sin }^2\theta =\frac{1-\cos 2\theta }{2}{\cos }^2\theta =\frac{1+\cos 2\theta }{2}$$
$${\tan }^2\theta =\frac{1-\cos 2\theta }{1+\cos 2\theta }{e}^{i\theta }=\cos \theta +i\sin \theta {\cosh }^2\theta -{\sinh }^2\theta =11-{\tanh }^2\theta =\frac{1}{{\cosh }^2\theta }\sinh (a\pm b)=\sinh a\cosh b\pm \cosh a\sinh b$$
$$\cosh (a\pm b)=\cosh a\cosh b\pm \sinh a\sinh b\tanh (a\pm b)=\frac{\tanh a\pm \tanh b}{1\pm \tanh a\tanh b}\sinh 2x=2\sinh x\cosh x\cosh 2x=2\cosh x^2-1\tanh 2x=\frac{2\tanh x}{1+{\tanh }^{2}x}$$
$$\sinh 3x=4\sinh x^3+3\sinh x\cosh 3x=4{\cosh }^{3}x-3\cosh x\tanh 3x=\frac{3\tanh x+{\tanh }^{3}x}{1+3{\tanh }^{2}x}\sinh x=-i\sin \left(ix\right)\cosh x=\cos \left(ix\right)$$
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2Ra=b\cos C+c\cos B{a}^2=b^2+c^2-2bc\cos A\cos A=\frac{b^2+c^2-a^2}{2bc}\frac{a-b}{a+b}=\frac{\tan \left(\frac{1}{2}(A-B)\right)}{\tan \left(\frac{1}{2}(A+B)\right)}\cot \left(\frac{A}{2}\right)=\frac{s-a}{\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}}$$
$$\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{1}{4}(\cos A+\cos B+\cos C-1)=\frac{(s-a)(s-b)(s-c)}{abc}{\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2(\cos A\cos B\cos C+1)$$
$$\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}=\frac{1}{4}(\sin A+\sin B+\sin C)=\frac{s}{4R}\tan A\tan B\tan C=\tan A+\tan B+\tan C\frac{1}{\tan A_{/2}\tan B_{/2}\tan C_{/2}}=\frac{1}{\tan A_{/2}}+\frac{1}{\tan B_{/2}}+\frac{1}{\tan C_{/2}}$$
$$r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}S=\frac{1}{2}bc\sin A=rs=\frac{abc}{4R}=2R^2\sin A\sin B\sin C=\frac{a^2\sin B\sin C}{2\sin A}=\sqrt{s(s-a)(s-b)(s-c)}=r_A(s-a)=\sqrt{r{r}_A{r}_B{r}_C}$$
$$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}{f}^{\prime }\left(a\right)=\underset{h\to 0}{lim}\frac{f(a+h)-f\left(a\right)}{h}{\left(fg\right)}^{\prime }=f^{\prime }g+f{g}^{\prime }{\left(fg\right)}^{″}=f^{″}g+2f^{\prime }{g}^{\prime }+f{g}^{″}{\left(fg\right)}^{\left(n\right)}=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{\left(k\right)}{g}^{(n-k)}{\left(\frac{1}{f}\right)}^{\prime }=-\frac{f^{\prime }}{f^2}$$
$${\left(\frac{g}{f}\right)}^{\prime }=\frac{g^{\prime }f-f^{\prime }g}{f^2}{(f\circ g)}^{\prime }=f^{\prime }\circ g{g}^{\prime }{\left(x^a\right)}^{\prime }=a{x}^{a-1}{(\sin x)}^{\prime }=\cos x{(\cos x)}^{\prime }=-\sin x{(\tan x)}^{\prime }=\frac{1}{{\cos }^{2}x}{(\cot x)}^{\prime }=-\frac{1}{{\sin }^{2}x}{\left(a^x\right)}^{\prime }=a^x\log a$$
$${({\log }_{a}x)}^{\prime }=\frac{1}{x\log a}{(\arcsin x)}^{\prime }=\frac{1}{\sqrt{1-x^2}}{(\arccos x)}^{\prime }=-\frac{1}{\sqrt{1-x^2}}{(\arctan x)}^{\prime }=\frac{1}{1+x^2}{(\sinh x)}^{\prime }=\cosh x{(\cosh x)}^{\prime }=\sinh x{(\tanh x)}^{\prime }=1-{\tanh }^{2}x$$
$${(\log x)}^{(n)}=-(n-1)!{(-x)}^{-n}\int {(x-a)}^{t}dx=\frac{1}{t+1}{(x-a)}^{t+1}+C{\int }_{\alpha }^{\beta }(x-\alpha )(\beta -x)dx=\frac{{(\beta -\alpha )}^3}{6}{\int }_a^{b}f\left(x\right)dx=\frac{b-a}{6}(f\left(a\right)+4f(\frac{a+b}{2})+f\left(b\right))$$
$${\int }_0^1{x}^m{(1-x)}^{n}dx=\frac{m!n!}{(m+n+1)!}\int \frac{1}{\sin x}dx=\frac{1}{2}\log \left(\frac{1-\cos x}{1+\cos x}\right)+C\int \frac{1}{\cos x}dx=\frac{1}{2}\log \left(\frac{1+\sin x}{1-\sin x}\right)+C\int \frac{1}{\tan x}dx=\log |\sin x|+C$$
$$\int e^{ax}\sin bxdx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)+C\int e^{ax}\cos bxdx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+C\int \frac{1}{\sqrt{x^2+a^2}}dx=\log (x+\sqrt{x^2+a^2})+C$$
$$\int \sqrt{x^2+a^2}dx=\frac{1}{2}(x\sqrt{x^2+a^2}+a^2\log (x+\sqrt{x^2+a^2}))+C\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+C{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a{x}^2}dx=\sqrt{\frac{\pi }{a}}{\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x-1}dx=\frac{{\pi }^4}{15}$$
$${\int }_0^{\mathrm{\infty }}{x}^{2n+1}{e}^{-a{x}^2}dx=\frac{n!}{2a^{n+1}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2n}{e}^{-a{x}^2}dx=\frac{(2n-1)!!}{2^n{a}^n}\sqrt{\frac{\pi }{a}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a{x}^2+bx+c}dx=\sqrt{\frac{\pi }{a}}{e}^{\frac{b^2}{4a}+c}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\sin }^{2}xdx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\cos }^{2}xdx=\sqrt{\frac{\pi }{2}}$$
$$S={\int }_a^b\sqrt{1+f^{\prime }{\left(x\right)}^2}dx={\int }_{\alpha }^{\beta }\frac{1}{2}(x{y}^{\prime }-y{x}^{\prime })dt={\int }_{\alpha }^{\beta }\frac{1}{2}r{\left(\theta \right)}^{2}d\theta L={\int }_{\alpha }^{\beta }\sqrt{f^{\prime }{\left(t\right)}^2+g^{\prime }{\left(t\right)}^2}dt={\int }_{\alpha }^{\beta }\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}d\theta$$
$$V={\int }_a^b\pi {\left(f\left(x\right)\right)}^{2}dx=\frac{2}{3}\pi {\int }_{\alpha }^{\beta }r{\left(\theta \right)}^3\sin \theta d\theta ={\int }_a^{b}2\pi x\left|f\left(x\right)\right|dx=\pi \cos \theta {\int }_a^b{(mx-f\left(x\right))}^{2}dx\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}e-tdt$$
$$\zeta \left(s\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}\sinh \left(x\right)=\frac{e^x-e^{-x}}{2}\cosh \left(x\right)=\frac{e^x+e^{-x}}{2}\tanh \left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}\mathrm{gd}\left(x\right)={\int }_0^x\frac{1}{\cosh t}dt{\varsigma }_a\left(x\right)=\frac{1}{1+e^{-ax}}=\frac{\tanh \left(\frac{ax}{2}\right)+1}{2}$$
$$F(x,k)={\int }_0^x\frac{1}{\sqrt{(1-t^2)(1-k^2{t}^2)}}dtE(x,k)={\int }_0^x\sqrt{\frac{1-k^2{t}^2}{1-t^2}}dt\mathrm{\Pi }(a;x,k)={\int }_0^x\frac{1}{(1-a{t}^2)\sqrt{(1-t^2)(1-k^2{t}^2)}}dt{J}_n\left(x\right)=\frac{1}{\pi i^n}{\int }_0^{\pi }{e}^{ix\cos \theta }\cos n\theta d\theta$$
$$\sum _{k=1}^{n}k=\frac{1}{2}n(n+1)\sum _{k=1}^n{k}^2=\frac{1}{6}n(n+1)(2n+1)\sum _{k=1}^n{k}^3={\left(\frac{1}{2}n(n+1)\right)}^2\sum _{k=1}^n{k}^a=\frac{1}{a+1}\sum _{k=0}^a(\genfrac{}{}{0ex}{}{a+1}{k})B_k{n}^{a-k+1}\prod _{k=1}^n{a}^{a_k}=a^{\sum _{k=1}^n{a}_k}$$
$$\prod _{k=1}^n(k+a)=\frac{(a+n)!}{a!}{F}_n=\frac{1}{\sqrt{5}}({\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n)V_n=\frac{{\pi }^{\frac{n}{2}}}{\mathrm{\Gamma }(\frac{n}{2}+1)}{e}^x=\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}\sin x=\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k\frac{x^{2k+1}}{(2k+1)!}\cos x=\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k\frac{x^{2k}}{(2k)!}$$
$$\tan x=\sum _{k=1}^{\mathrm{\infty }}\frac{B_{2k}{(-4)}^k(1-4^k)}{\left(2k\right)!}{x}^{2k-1}\log (x+1)=\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k\frac{x^{k+1}}{k+1}\frac{1}{{(1-ax)}^m}=\sum _{k=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m+k-1}{m-1})a^k{x}^k{W}_0\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{(-k)}^{k-1}}{k!}{x}^k$$
$$\sinh x=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(2k+1)!}{x}^{2k+1}\cosh x=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{\left(2k\right)!}{x}^{2k}\tanh x=\sum _{k=1}^{\mathrm{\infty }}\frac{B_{2k}{4}^k(4^k-1)}{\left(2k\right)!}{x}^{2k-1}\sum _{k=1}^{\mathrm{\infty }}k=-\frac{1}{12}\sum _{k=1}^{\mathrm{\infty }}{F}_k=-1\sum _{k=0}^n(\genfrac{}{}{0ex}{}{p}{k})(\genfrac{}{}{0ex}{}{q}{n-k})=(\genfrac{}{}{0ex}{}{p+q}{n})$$
$$R{i}_n=(\genfrac{}{}{0ex}{}{n}{4})+\frac{-5n^3+45n^2-70n+24}{24}{\delta }_2\left(n\right)-\frac{3n}{2}{\delta }_4\left(n\right)+\frac{-45n^2+262n}{6}{\delta }_6\left(n\right)+42n{\delta }_{12}\left(n\right)+60n{\delta }_{18}\left(n\right)+35n{\delta }_{24}\left(n\right)-38n{\delta }_{30}\left(n\right)-82n{\delta }_{42}\left(n\right)-330n{\delta }_{60}\left(n\right)-144n{\delta }_{84}\left(n\right)-96n{\delta }_{90}\left(n\right)-144n{\delta }_{120}\left(n\right)-96n{\delta }_{210}\left(n\right)$$
$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}=\prod _{p:prime}\frac{1}{1-\frac{1}{p^s}}\sum _{n=1}^{\mathrm{\infty }}\frac{\lambda \left(n\right)}{n^s}=\frac{\zeta \left(2s\right)}{\zeta \left(s\right)}\sum _{n=1}^{\mathrm{\infty }}\frac{\mu \left(n\right)}{n^s}=\frac{1}{\zeta \left(s\right)}\pi =4\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}=2\sqrt{3}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{3^n(2n+1)}=\prod _{n=1}^{\mathrm{\infty }}(\frac{2n}{2n-1}\cdot \frac{2n}{2n+1})$$
$$\frac{1}{\pi }=\frac{2\sqrt{2}}{{99}^2}\sum _{n=0}^{\mathrm{\infty }}\frac{(26390n+1103)\cdot \left(4n\right)!}{{(4^n{99}^n\cdot n!)}^4}=\frac{12}{640320\sqrt{640320}}\sum _{n=0}^{\mathrm{\infty }}\frac{(545140134n+13591409)\cdot \left(6n\right)!}{{(-640320)}^{3n}\cdot \left(3n\right)!\cdot {(n!)}^3}\frac{4}{\pi }=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n(21460n+1123)\cdot \left(4n\right)!}{{882}^{2n+1}{(4^{n}n!)}^4}$$
$$\frac{1}{\pi }=\frac{12}{(1249638720+159999840\sqrt{61})\sqrt{1249638720+159999840\sqrt{61}}}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(6n)!(1657145277365+212175710912\sqrt{61}+(107578229802750+3773980892672\sqrt{61})n)}{(3n)!(n!{)}^3{(1249638720+159999840\sqrt{61})}^n}$$