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Ta có TQ: (phân số đầu - phân số cuối) : khoảng cách
Áp dụng vào bài toán => (\(\frac{1}{1}\)-\(\frac{1}{100}\)) : 1 =\(\frac{99}{100}\)
lý dó 1 là khoảng cách vì cách lm như sau: 2-1=1
3-2=1
.....
100-99=1
=> khoảng cách là 1
Chúc bn hk tốt nhé!!
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{5}{4.5}-\frac{4}{4.5}+...+\frac{99}{98.99}-\frac{98}{98.99}+\frac{100}{99.100}-\frac{99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
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=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
...
= 1/2-1/3+1/3-1/4+...+ 1/19-1/20
= 1/2-1/20
=9/20
có phải như thế này ko bn
\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}\)
A = \(\frac{9}{20}\)
\(B=\frac{1}{99.100}-\frac{1}{98.99}-\frac{1}{97.98}-.....-\frac{1}{1.2}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
\(B=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)\)
B = \(-\frac{99}{100}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
= \(1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{98}-\frac{1}{98}\right)-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
Vậy ...
B = \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
= \(\frac{1}{2}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{17}-\frac{1}{17}\right)-\frac{1}{20}\)
= \(\frac{1}{2}-\frac{1}{20}\)
= \(\frac{9}{20}\)
Vậy B = 9/20
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy......
Ta có \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{249.250}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{249}-\frac{1}{250}\)
\(=1-\frac{1}{250}\)
\(=\frac{249}{250}\)
a) = 1-1/2+1/2-1/3+1/3-1/4
= 1-1/4=3/4
b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018
=1-1/2018=2017/2018
c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015
= 1/2-1/2015=2015/4030-2/4030=2013/4030
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)
\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}.\frac{2013}{4030}\)
\(=\frac{6039}{8060}\)
Câu hỏi của lương hiếu - Toán lớp 6 - Học toán với OnlineMath
Làm như link trên nhưng bỏ hạng tử \(\frac{1}{99.100}\)đi
Bước cuối: \(1-\frac{1}{99}=\frac{98}{99}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)
\(\Rightarrow C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow C=1-\frac{1}{99}\)
\(\Rightarrow C=\frac{98}{99}\)
~Study well~
#KSJ
=1-1/2 +1/2-1/3 +1/3-1/4 +....+1/97.98+1/98+1/99
=1-1/99
=98/99
1/97-1/98 nha
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(\Rightarrow C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow C=1-\frac{1}{99}\)
\(\Rightarrow C=\frac{98}{99}\)
rất vui vì giúp đc bạn !!!
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}\)
\(=1-\frac{1}{98}\)
\(=\frac{98}{98}-\frac{1}{98}\)
\(=\frac{97}{98}\)
\(C=\frac{1}{1.2} +\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{97.98}+\frac{1}{98.99}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)
\(=\frac{1}{1}-\frac{1}{99}\)
\(=\frac{98}{99}\)