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HT
9 tháng 11 2020
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}\)
\(=\frac{5}{6}\)
LM
Lê Minh Vũ
CTVHS
13 tháng 7 2021
\(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\)\(\frac{1}{5.6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{5}{6}\)
Hok tốt
9 tháng 5 2017
Ta có :
\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+..............+\dfrac{1}{99.100}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...........+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S=1-\dfrac{1}{100}=\dfrac{99}{100}\)
LA
10 tháng 8 2019
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
DN
2
24 tháng 4 2019
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
24 tháng 4 2019
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\)
=>\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
=> 1-\(\frac{1}{6}\)
=\(\frac{6}{6}-\frac{1}{6}=\frac{6}{6}+\frac{-1}{6}=\frac{5}{6}\)
NH
5
I
4 tháng 6 2021
A=12/1.2 .22/2.3 .32/3.4 .42/4.5
=1/2. 2.2/2.3 .3.3/3.4 .4.4/4.5
=1/2.2/3.3.4.4./5
=1/5
18 tháng 7 2016
c)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
18 tháng 7 2016
\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)
\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=1\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
Lời giải
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)