Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{189}{190}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}........\frac{378}{380}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}........\frac{18.21}{19.20}=\frac{1.2.3......18}{2.3.4....19}.\frac{4.5.6....21}{3.4.5....20}\)
\(P=\frac{1}{19}.\frac{21}{3}=\frac{21}{57}\)
\(A=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right).....\left(1-\frac{1}{1326}\right)\)
\(A=\frac{20}{21}.\frac{27}{28}.\frac{35}{36}......\frac{1325}{1326}\)
\(A=\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}.....\frac{50.53}{51.52}\)
\(A=\frac{5.\left(6.7.....50\right)}{\left(6.7.....50\right).51}.\frac{53.\left(8.9.10.....52\right)}{7.\left(8.9.10.......52\right)}\)
\(A=\frac{5}{51}.\frac{53}{7}=\frac{265}{357}\)
\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
\(\Rightarrow E=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)
\(=1+\frac{1}{2}\left(3+4+5+...+201\right)\)
\(=1+\frac{1}{2}\left(1+2+3+...+201-1-2\right)\)
\(=1+\frac{1}{2}\left(\frac{201.202}{2}-3\right)=10150\)
\(\frac{21}{5}\left|x\right|< 2019\Rightarrow\left|x\right|< 2019\div\frac{21}{5}=\frac{3365}{7}\)
\(\Rightarrow-480\le x\le480\)
\(\Rightarrow\sum x=-480+480-479+479+...+-1+1+0=0\)
\(\frac{2^{24}\left(x-3\right)}{\frac{81}{35}.\left(6.2^{24}-2^{26}\right)}=\frac{25}{9}\)
\(\Leftrightarrow\frac{2^{24}\left(x-3\right)}{2^{24}\left(6-2^2\right)}=\frac{25}{9}.\frac{81}{35}\)
\(\Leftrightarrow\frac{x-3}{2}=\frac{45}{7}\)
\(\Leftrightarrow x-3=\frac{90}{7}\)
\(\Rightarrow x=\frac{111}{7}\)
\(\left(\frac{1}{7}x-\frac{1}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)
\(\orbr{\begin{cases}\frac{1}{7}x-\frac{1}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{7}{3}\\x=3\end{cases}}\)
\(\left(\frac{1}{7}x-\frac{1}{3}\right).\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)
=>\(\frac{1}{7}x-\frac{1}{3}=0=>\frac{1}{7}x=\frac{1}{3}=>x=\frac{7}{3}\)
\(-\frac{1}{5}x+\frac{3}{5}=0=>-\frac{1}{5}x=-\frac{3}{5}\) =>\(x=3\)
ta có
\((1-\frac{1}{2})\times(1-\frac{1}{3})\times(1-\frac{1}{4})\times...(1-\frac{1}{n-1})\times(1-\frac{1}{n})\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{n-2}{n-1}\times\frac{n-1}{n}\)
\(=\frac{1\times2\times3\times...\times(n-2)\times(n-1)}{2\times3\times4\times...\times(n-1)\times n}\)
\(=\frac{1}{n}\)