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b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+........+\frac{1}{99\cdot100}\right)-2x=\frac{1}{2}\)
\(\left(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{100-99}{99\cdot100}\right)-2x=\frac{1}{2}\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\)
\(\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\)
\(\frac{99}{100}-2x=\frac{1}{2}\)
\(2x=\frac{99}{100}-\frac{1}{2}\)
\(2x=\frac{49}{100}\)
\(x=\frac{49}{100}:2\)
\(x=\frac{49}{200}\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\)
\(\frac{99}{100}-2x=\frac{1}{2}\)
\(\frac{99-50}{100}=2x\)
\(49=200x\)
\(x=\frac{49}{200}\)
Ta có :
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)
\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)
\(=\frac{2015}{1}.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)
Vậy \(x=2015\)
Ủng hộ mk nha !!! ^_^
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}\)
\(x=\frac{23}{11}\)
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{x\left(x+1\right)}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{215}{216}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{216}\)
\(\Leftrightarrow x=216-1=215\)
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
a/(x+1)+(x+2)+...+(x+100)=5750
x+1+x+2+...+x+100=5750
100x+(1+2+...+100)=5750
100x+50.(100+1)=5750
100x+5050=5750
100x=700
x=7.
b/ 1/1.2+1/2.3+...+1/x(x+1)=2015/2016
1/1-1/2+1/3-1/4+...+1/x-1/x+1=2015/2016
1-1/x+1=2015/2016
1/x+1=1/2016
x+1=2016
x=2015