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`A=1/2+1/6+1/12+1/20+1/30+...+1/9900`
`=1/(1xx2)+1/(2xx3)+1/(3xx4)+1/(4xx5)+1/(5xx6)+...+1/(99xx100)`
`=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100`
`=1/1-1/100`
`=100/100-1/100`
`=99/100`
ta có:
1/2+1/6+...+1/9900
=1/1.2+1/2.3...+1/99.100
=1-1/2+1/2-1/3+1/3-...+1/99-1/100
=1-1/100
=99/100
\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{9900}\)
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\cdot\cdot\cdot+\frac{1}{99\times100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=1-\dfrac{1}{6}=\dfrac{5}{6}\)
a×1/2+a×1/6+a×1/12
vơi a =8
\(\Rightarrow\)8×1/2+8×1/6+8×1/12=8×(1/2×1/6+1/12)=8×3/4=6
T= 1 - 1/2 + 1/2 - 1/3 + ......+ 1/99 - 1/100
= 1 - 1/100
= 99/100
\(t=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(t=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(t=1-\frac{1}{100}=\frac{99}{100}\)
Vậy \(t=\frac{99}{100}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{1}{1}-\dfrac{1}{100}\)
\(A=\dfrac{99}{100}\)
\(\cdot\) LÀ DẤU \(\times\)
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+.....+ \(\dfrac{1}{9900}\)
A = \(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+....+\dfrac{1}{99\times100}\)
A = \(\dfrac{1}{1}-\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)+......+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
A = \(\dfrac{99}{100}\)
ta có : t = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/98.99 + 1/99.100
=> t = 1/1 - 1/2 + 1/2 - 1/3 + .... + 1/99 - 1/100
=> t = 1 - 1/100
=> t = 99/100
T=1/1x2+1/2x3+1/3x4+....................+1/98x99+1/99x100
T=1-1/2+1/2-1/3+..............+1/98-1/99+1/99-1/100
T=1-1/100
T=99/100
\(\dfrac{1}{2}\times\dfrac{3}{4}:\dfrac{6}{5}=\dfrac{3}{8}:\dfrac{6}{5}=\dfrac{5}{16}\)
\(\dfrac{2}{3}+\dfrac{1}{6}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{14}{21}+\dfrac{6}{21}=\dfrac{20}{21}\)
\(8\times\dfrac{3}{5}:\dfrac{12}{5}=\dfrac{24}{5}:\dfrac{12}{5}=2\)
\(\dfrac{4}{9}+\dfrac{3}{4}-\dfrac{1}{3}=\dfrac{16}{36}+\dfrac{27}{36}-\dfrac{12}{36}=\dfrac{31}{36}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{100}\Rightarrow A=\frac{99}{100}\)
\(\text{Vậy }A=\frac{99}{100}\)
\(A=\)\(\frac{1}{2}\)\(+\)\(\frac{1}{6}\)\(+\)\(\frac{1}{12}\)\(+\)\(\frac{1}{20}\)\(+\)\(\frac{1}{30}\)\(+...+\)\(\frac{1}{9900}\)
\(A=\)\(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\)\(\frac{1}{5.6}\)\(+...+\)\(\frac{1}{99.100}\)
\(A=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{6}\)\(+...+\)\(\frac{1}{99}\)\(-\)\(\frac{1}{100}\)
\(A=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{100}\)
\(A=\)\(\frac{99}{100}\)