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kết quả phép tính
1/1x2+1/2x3+...................+1/2015x2016+1/2016
A:1/2016 B:2015/2016 C:1/2015 D:1
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\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2015\times2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}+\frac{1}{2016\cdot2017}\)
\(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{2016-2015}{2015\cdot2016}+\frac{2017-2016}{2016\cdot2017}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)(làm gọn một chút)
\(1-\frac{1}{2017}=\frac{2016}{2017}\)
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\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{2015x2016}+\frac{1}{2016}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1015}-\frac{1}{2016}+\frac{1}{2016}=1\)
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1/(1×2) + 1/(2×3) + 1/(3×4) + ... + 1/(2021×2022)
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2021 - 1/2022
= 1 - 1/2022
= 2021/2022
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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 = \(\dfrac{2021}{2022}\)
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\(a,\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
a)=(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/99-1/100)
=1-1/2+1/2-1/3+1/3-1/4+......+1/99-1/100
=1-1/100=99/100
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`x/(x+1)=1/(1xx2)+1/(2xx3)+1/(3xx4)+...+1/(31xx32)`
`=>x/(x+1)=1-1/2+1/2-1/3+1/3-1/4+...+1/31-1/32`
`=>x/(x+1)=1-1/32`
`=>x/(x+1)=31/32`
`=>32x=31(x+1)`
`=>32x=31x+31`
`=>32x-31x=31`
`=>x=31`
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{14.15}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{14}-\frac{1}{15}\)
\(A=1-\frac{1}{15}\)
\(A=\frac{14}{ }15\)
đề kiểu gì thế, ai mà làm được