Tính bằng cách hợp lý nhất: 1/1x2 + 1/2x3+1/3x4+.....+1/998x999+1/999x1000
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Ta có \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{998\times999}+\frac{1}{999\times1000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}\)
\(=\frac{999}{1000}\)
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
=1- 1/2 +1/2- 1/3+....+1/9- 1/10
=1- 1/10
=9/10
1/1x2 + 1/2x3 + 1/3x4 +...+1/9x10 = 1/1 - 1/2 + 1/2 -1/3 + 1/3 -1/4 +...+1/9-1/10
=1+( 1/2 -1/2)+(1/3-1/3)+...+(1/9-1/9)-1/10
=1 + 0 + 0 +...-1/10 = 9/10
A=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
A=\(\frac{1}{2}-\frac{1}{7}\)
A=\(\frac{5}{14}\)
A = 1/2 -1/3 +1/3-1/4 + 1/4-1/5 +1/5-1/6 + 1/6-1/7 =
1/2-1/7 = 5/14
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
`A=1/(2xx3)+1/(3xx4)+1/(4xx5)+1/(5xx6)+1/(6xx7)+1/(7xx8)`
`=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8`
`=1/2-1/8`
`=4/8-1/8`
`=3/8`
Vậy `A=3/8`
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)
b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)
c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)