Tính bằng cách hợp lý nhất: 1/1x2 + 1/2x3+1/3x4+.....+1/998x999+1/999x1000
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\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
=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 = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+...+ \(\dfrac{1}{2021\times2022}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+...+ \(\dfrac{1}{2021}\) - \(\dfrac{1}{2022}\)
A = 1 - \(\dfrac{1}{2022}\)
A = \(\dfrac{2021}{2022}\)
\(\frac{1}{1x2}+\frac{1}{1x3}+...+\frac{1}{999x1000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}\)
\(=\frac{999}{1000}\)
1/1x2+1/2x3+1/3x4+...+1/999x1000
=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000
=1-1/1000
=1000/1000-1/1000
=999/1000
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
\(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+...+\(\dfrac{1}{99\times100}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
Ta có \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{998\times999}+\frac{1}{999\times1000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}\)
\(=\frac{999}{1000}\)