Tìm chữ số tận cùng của $4^{736}$ |x-1|+|x-2|+|x-3|$$ |x+2|+|x+3|$$ \begin{array}{c}C=\frac{3}{2}+{\left(\frac{3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^3+....+{\left(\frac{3}{2}\right)}^{2013}\\ \iff \frac{3}{2}C=\frac{3}{2}.\left[\frac{3}{2}+{\left(\frac{3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^3+....+{\left(\frac{3}{2}\right)}^{2013}\right]\\ \iff \frac{3}{2}C={\left(\frac{3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^3+{\left(\frac{3}{2}\right)}^4+......+{\left(\frac{3}{2}\right)}^{2014}\\ \iff \frac{3}{2}C-C=\left[{\left(\frac{3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^3+{\left(\frac{3}{2}\right)}^4+......+{\left(\frac{3}{2}\right)}^{2014}\right]-\left[\frac{3}{2}+{\left(\frac{3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^3+....+{\left(\frac{3}{2}\right)}^{2013}\right]\\ \iff \frac{1}{2}C={\left(\frac{3}{2}\right)}^{2014}-\left(\frac{3}{2}\right)\\ \iff C={\left(\frac{3}{2}\right)}^{2014}.2-3\\ A=\frac{1}{2}+C={\left(\frac{3}{2}\right)}^{2014}.2-\frac{5}{2}\\ B-A=\left({\left(\frac{3}{2}\right)}^{2013}.2\right)-\left({\left(\frac{3}{2}\right)}^{2014}.2-\frac{5}{2}\right)\\ =2.{\left(\frac{3}{2}\right)}^{2013}.\left(1-\frac{3}{2}\right)+\frac{5}{2}\\ =-{\left(\frac{3}{2}\right)}^{2013}+\frac{5}{2}\end{array}$$ M⋮7;M⋮8;M⋮50$$ (x-1{)}^2$$ (y-1{)}^2$$ x-5{)}^8$$ x-5{)}^3$$ \frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3n-1\right).\left(3n+2\right)}=\frac{n}{6n+4}$$ \frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}=\frac{5n}{4n+3}$$ \frac{8}{3}+\frac{7}{8^2}\frac{5}{8^3}\frac{32}{8^5}vàB=\frac{24}{8^2}+40+\frac{5}{8^2}+\frac{40}{8^2}\frac{5}{8^4}.$$ \frac{1}{5}$$ +$$ \frac{1}{8}$$ +$$ \frac{1}{11}$$ +$$ \frac{1}{14}$$ +$$ +$$ \frac{1}{92}$$ +$$ \frac{1}{95}$$ a,b,c$$ a+b+c=3$$ P=a{b}^2+b{c}^2+c{a}^2$x^2{y}^2{y}^2$ \in $$ \frac{x}{15}=\frac{4}{y}=\frac{-2}{5}$$ \frac{-5}{9}+\frac{-8}{15}+\frac{22}{-9}+\frac{-7}{15}<x\le \frac{-1}{3}+\frac{-1}{4}+\frac{-5}{12}$$ \frac{x-3}{4}=\frac{15}{20}$$