=> S = \(\frac{1}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+....+\frac{1}{97.100}\right)\)
= \(\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\)
= \(\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
\(S=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
1-1/4cong 1/4 -1/7 cong.....cong 1/97-1/100
1-1/100
99/100
S= \(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+,,,+ \(\frac{1}{97.100}\)
\(\Rightarrow\)S =\(\frac{1}{3}\)(1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{7}\)+...+ \(\frac{1}{97}\)- \(\frac{1}{100}\))
=\(\frac{1}{3}\). (1 - \(\frac{1}{100}\))
=\(\frac{1}{3}\). \(\frac{99}{100}\)
=\(\frac{33}{100}\)
mk có cách khác nè :
Ta có : \(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}\)
\(\Rightarrow3S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)
\(\Rightarrow3S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3S=1-\frac{1}{100}\)
\(\Rightarrow3S=\frac{99}{100}\)
\(\Rightarrow S=\frac{99}{100}.\frac{1}{3}=\frac{33}{100}\)
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
S= 1/1.4+1/4.7+1/7.10+...+1/97.100
suy ra S*3=3*(1/1.4+1/4.7+....+1/97.100)
s*3=3/1.4+3/4.7+....+3/97*100
s*3=4-1/1.4+7-4/4.7+....+100-97/97.100
s*3=1/1-1/100
s*3=99/100
s=99/100:3
s=.....
vậy s=.....
\(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(S=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\frac{99}{100}\)
\(S=\frac{99}{300}=\frac{33}{100}\)
\(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(< =>3S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(< =>3S=1-\frac{1}{100}=\frac{99}{100}\)
\(< =>S=\frac{99}{100}:3=\frac{33}{100}\)
\(S=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\)
\(3S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\)
\(3S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}+\frac{1}{100}\)
\(3S=1-\frac{1}{100}\)
\(3S=\frac{99}{100}\)
\(S=\frac{33}{100}\)
S = \(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+ .....+\(\frac{1}{97.100}\)
= \(\frac{1}{3}\)\((\frac{1}{1}\)-\(\frac{1}{4}\)+\(\frac{1}{4}-\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{10}\)+ ....+\(\frac{1}{97}\)-\(\frac{1}{100})\)
= \(\frac{1}{3}\)\((1-\frac{1}{100}\)\()\)
= \(\frac{1}{3}\) . \(\frac{99}{100}\)= \(\frac{33}{100}\)
hok tốt
Bài làm :
Ta có :
\(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{97.100}\)
\(\Rightarrow3S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
\(\Leftrightarrow3S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(\Leftrightarrow3S=1-\frac{1}{100}\)
\(\Leftrightarrow3S=\frac{99}{100}\)
\(\Leftrightarrow S=\frac{99}{100}\div3=\frac{33}{100}\)
Vậy S=33/100
