\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
\(A=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\cdot\cdot\cdot\cdot\frac{9^2}{9\cdot10}\)
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}=\frac{1^2.2^2.3^2...9^2}{1.2.2.3.3.4.4...9.10}=\frac{1.2^2.3^2...9^2}{1.2^2.3^2.4^2...10^2}=\frac{1}{10^2}=\frac{1}{100}\)
\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)
\(\Rightarrow A=B\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+..+\frac{1}{8\cdot9\cdot10}\cdot x=\frac{22}{45}\)thì x=
B cho mình hỏi hình như thiếu dấu ngoặc
Tìm x biết: \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\cdot x=\frac{22}{45}\)Trình bày cách tính
X=2 nhé bạn.....đúng đó, vòng 12 mk 300 mà cx gặp câu này!!! Tick nha
\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^4}{3\cdot5}\cdot\cdot\cdot\frac{50^2}{49\cdot51}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{50^2}{49.51}=\frac{\left(1.2.3.4...50\right)^2}{1.2.3.4...50.51}=\frac{1.2.3...50}{51}=\frac{50!}{51}\)
\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{50^2}{49\cdot51}\)
\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\frac{5^2}{4\cdot6}\cdot\frac{7^2}{5\cdot7}\cdot\cdot\cdot\frac{50^2}{49\cdot51}\)
\(=\frac{2}{1}\cdot\frac{50}{51}=\frac{100}{51}\)
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{50^2}{49.51}\)
\(=\frac{2^2.3^2.4^2...49^2.50^2}{1.3.2.4.3.5...48.50.49.51}\)
\(=\frac{2.50}{1}=100\)
tính \(A=\)\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\cdot...\cdot\frac{99^2}{99\cdot100}\)
A = \(\frac{\left(1.2.3......99\right)\left(1.2.3......99\right)}{\left(1.2.3......99\right)\left(2.3.4.....100\right)}=\frac{1.1}{1.100}=\frac{1}{100}\)
Tính B=\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{2015^2}{2014\cdot2016}\)
\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}...\frac{2015.2015}{2014.2016}\)
\(B=\frac{2.3...2015}{1.2...2014}.\frac{2.3...2015}{3.4...2016}\)
\(B=2015.\frac{1}{1008}\)
\(B=\frac{2015}{1008}\)
G=\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{50^2}{49.51}\)
H=\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{10}{7}\right)\)
Giúp mình vs
G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)
\(G=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)
\(=\frac{\left(2.3.4.....50\right).\left(2.3.4.....50\right)}{\left(1.2.3.....49\right).\left(3.4.5.....51\right)}\)
\(=\frac{50.2}{51}=\frac{100}{51}\)
\(H=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{7}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....0.....\left(1-\frac{10}{7}\right)\)
\(=0\)
Cúc Nguyễn
