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= \(\frac{1}{4}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{19}\)+... + \(\frac{1}{44}\)-\(\frac{1}{49}\)
= \(\frac{1}{4}\)-\(\frac{1}{49}\)
= \(\frac{45}{196}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm nè huhu
5B=\(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+...+\frac{5}{64\cdot69}\)
5B=\(\frac{9-4}{4\cdot9}+\frac{14-9}{9\cdot14}+...+\frac{69-64}{64.69}\)
5B=\(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)
5B=\(\frac{65}{276}\)
B=\(\frac{13}{276}\)
\(B=\frac{1}{4.9}+\frac{1}{9.14}+....+\frac{1}{64.69}\)
\(\Rightarrow5B=\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{64.69}\)
\(5B=\frac{9-4}{4.9}+\frac{14-9}{9.14}+....+\frac{69-64}{64.69}\)
\(5B=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)
\(5B=\frac{1}{4}-\frac{1}{69}\)
\(5B=\frac{65}{276}\)
\(B=\frac{65}{276}:5\)
\(B=\frac{13}{276}\)
mk chỉ tiềm đc bài i hệt bài của bn
https://olm.vn/hoi-dap/detail/99402078680.html
\(D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2017}{2018}\)
\(D=\frac{1}{2018}\)
Vậy \(D=\frac{1}{2018}\)
\(E=\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{64.69}+\frac{1}{69.74}\)
\(E=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{69}-\frac{1}{74}\right)\)
\(E=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{74}\right)\)
\(E=\frac{1}{5}\cdot\frac{35}{148}=\frac{7}{148}\)
Vậy E = ...
a 2^2015>3^1029
b 5A=\(\frac{5}{4.9}\)+\(\frac{5}{9.14}\)+\(\frac{5}{14.19}\)+.....+\(\frac{5}{64.69}\)
5A=1/4-1/9+1/9-1/14+1/14-1/19+1/19+....+1/64-1/69
5A=1/4-1/9
A=(1/4-1/9)/5
A=1/36
\(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{49}+1\right)\)
= \(\left(\frac{1}{2}+\frac{2}{2}\right).\left(\frac{1}{3}+\frac{3}{3}\right).\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{49}+\frac{49}{49}\right)\)
= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{50}{49}\)
=\(\frac{3.4.5...50}{2.3.4...49}\)
=\(\frac{50}{2}\)
=25
