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\(A=\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)
\(A=4\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)
\(A=4\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=4\left(\frac{1}{1}-\frac{1}{2015}\right)\)
\(A=4\left(\frac{2015-1}{2015}\right)\)
\(A=4.\frac{2014}{2015}\)
... BẠN TỰ LÀM NỐT NHÉ!
Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\); \(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(S=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); \(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)
=> \(S< \frac{3}{4}\)
A= \(\frac{1^2}{1.2}\). \(\frac{2^2}{2.3}\). \(\frac{3^2}{3.4}\). \(\frac{4^2}{5.6}\).
A= \(\frac{1.1}{1.2}\). \(\frac{2.2}{2.3}\). \(\frac{3.3}{3.4}\). \(\frac{4.4}{4.5}\).
A= \(\frac{1.2.3.4}{1.2.3.4}\). \(\frac{1.2.3.4}{2.3.4.5}\).
A= 1. \(\frac{1}{5}\).
A= \(\frac{1}{5}\).
Vậy A= \(\frac{1}{5}\).
B= \(\frac{3}{4}\). \(\frac{8}{9}\). \(\frac{15}{16}\)..... \(\frac{899}{900}\).
B= \(\frac{1.3}{2.2}\). \(\frac{2.4}{3.3}\). \(\frac{3.5}{4.4}\)..... \(\frac{29.31}{30.30}\).
B= \(\frac{1.2.3.....29}{2.3.4.....30}\). \(\frac{3.4.5.....31}{2.3.4.....30}\).
B= \(\frac{1}{30}\). \(\frac{31}{2}\).
B= \(\frac{31}{60}\).
Vậy B= \(\frac{31}{60}\).
a) \(I=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)
b) \(K=\frac{4}{2\cdot4}+\frac{4}{2\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\)
\(\frac{1}{2}K=\frac{1}{2}\left(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\right)\)
\(\frac{1}{2}K=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\)
\(\frac{1}{2}K=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{2}{2010}\)
\(\frac{1}{2}K=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(K=\frac{2009}{2010}:\frac{1}{2}=\frac{2009}{1005}\)
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{4}{5}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{5}\)
\(1-\frac{1}{x+1}=\frac{4}{5}\)
\(\frac{x}{x+1}=\frac{4}{5}\)
\(\frac{x}{x+1}=\frac{4}{4+1}\)
\(\Rightarrow x=4\)
Vậy x = 4
=))
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{4}{5}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{5}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{x-1}=1-\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{x-1}=\frac{1}{5}\)
\(\Leftrightarrow x-1=5\)
\(\Leftrightarrow x=5+1\)
\(\Leftrightarrow x=6\)
~ Rất vui vì giúp đc bn ~ ^_<
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=1.\left(1-\frac{1}{101}\right)\)
\(B=\frac{100}{101}\)
\(C=\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{73.76}\)
\(C=4.\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)
\(C=4.\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
\(C=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(C=\frac{4}{3}.\frac{9}{38}\)
\(C=\frac{6}{19}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\\ =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\\ =1-\frac{1}{100}\\ =\frac{99}{100}\\ B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\\ =\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\\ =\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{4}{4.7}+\frac{4}{7.10}+....+\frac{4}{73.76}\\ =\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{73.76}\right)\\ =\frac{4}{3}.\left(\frac{3}{4}-\frac{3}{76}\right)\\ =\frac{18}{19}\)
Học tốt Nghe!!
Ta có: \(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)
\(=1\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2014.2015}\right)\)
\(=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=1\left(1-\frac{1}{2015}\right)\)
\(=1\left(\frac{2015}{2015}-\frac{1}{2015}\right)-1\left(\frac{2014}{2015}\right)=\frac{2014}{2015}\)
Vậy.....
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+....+\frac{4}{2014.2015}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\)
\(A=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)