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a, M=1/1.2+1/2.3+...+1/49.50
M=1−1/2+1/2−1/3+...+1/49−1/50
M=1−1/50<1
Vậy M<1
\(a,\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}< 1\)
\(=>M< 1\)
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
Tìm x :
a) \(2.x.\frac{-3}{4}=-\frac{5}{12}\)
\(\Rightarrow2x=-\frac{5}{12}:-\frac{3}{4}\)
\(\Rightarrow2x=\frac{5}{9}\)
\(\Rightarrow x=\frac{5}{9}:2\)
\(\Rightarrow x=\frac{5}{18}\)
Vậy : \(x=\frac{5}{18}\)
b) \(\frac{2}{3}+\frac{1}{3}.x=7\)
\(\Rightarrow\frac{1}{3}.x=7-\frac{2}{3}\)
\(\Rightarrow\frac{1}{3}.x=\frac{19}{3}\)
\(\Rightarrow x=\frac{19}{3}:\frac{1}{3}\)
\(\Rightarrow x=19\)
Vậy : \(x=19\)
c) \(\left(4.x+\frac{1}{8}\right)=\frac{3}{10}\)
\(\Rightarrow4.x=\frac{3}{10}-\frac{1}{8}\)
\(\Rightarrow4.x=\frac{7}{40}\)
\(\Rightarrow x=\frac{7}{40}:4\)
\(\Rightarrow x=\frac{7}{160}\)
Vậy : \(x=\frac{7}{160}\)
d) \(\frac{1}{3}.x-5=1\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}.x-5=\frac{3}{2}\)
\(\Rightarrow\frac{1}{3}.x=\frac{3}{2}+5\)
\(\Rightarrow\frac{1}{3}.x=\frac{13}{2}\)
\(\Rightarrow x=\frac{13}{2}:\frac{1}{3}\)
\(\Rightarrow x=\frac{39}{2}\)
Vậy : \(x=\frac{39}{2}\)
e) \(-\frac{2}{3}.x+\frac{1}{3}=-\frac{1}{2}\)
\(\Rightarrow-\frac{2}{3}.x=-\frac{1}{2}-\frac{1}{3}\)
\(\Rightarrow-\frac{2}{3}.x=-\frac{5}{6}\)
\(\Rightarrow x=-\frac{5}{6}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=\frac{5}{4}\)
Vậy : \(x=\frac{5}{4}\)
a) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)
b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)
c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)
d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)
e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)
Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2016}{2017}\)
\(=\frac{1.2.3......2016}{2.3.4.......2017}\)
\(=\frac{1}{2017}\)
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
\(M=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right).2.3.4...2018\)
\(\Rightarrow M=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right).2.3.4...673.674...2018\)
Vì \(\hept{\begin{cases}M⋮3\\M⋮673\end{cases}}\) mà \(\left(3,673\right)=1\) nên \(M⋮2019\left(đpcm\right)\)
\(M=\left[\left(1+\frac{1}{2018}\right)+\left(\frac{1}{2}+\frac{1}{2017}\right)+...+\left(\frac{1}{1008}+\frac{1}{1011}\right)+\left(\frac{1}{1009}+\frac{1}{1010}\right)\right].\)\(2.3...1008.1009.1010.1011...2017.2018\)
\(=\left(\frac{2019}{2018}+\frac{2019}{2.2017}+...+\frac{2019}{1008.1011}+\frac{2019}{1009.1010}\right).2.3...1008.1009.1010.1011...2017.2018\)
\(=2019\left(\frac{1}{2018}+\frac{1}{2.2017}+...+\frac{1}{1008.1011}+\frac{1}{1009.1010}\right).2...1008.1009.1010.1011...2017.2018\)
\(=2019.\left(2...2017+3...2016.2018+...+2.3...1007.1009.1011...2018+2.3....1008.1011...2018\right)\)
Chia hết cho 2019