tính
51.2:3.2-4.3 *[3-2.1]-2.68
Những câu hỏi liên quan
51.2 : 3.2 - 4.3 × ( 3 _ 2.1) - 2.68 =
Chỉ cho mình gấp nhé
Cảm ơn
tính nhanh : 66.65-65.64+64.63-63.62+....+4.3-3.2+2.1
tính P=-1+1/2.1+1/3.2+1/4.3+..........+1/2018.2017+1/2018
\(P=\)\(-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)
\(P=-1+1-\frac{1}{2018}+\frac{1}{2018}\)
\(P=0\)
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\(P=-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)
\(P=-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)
P = 0
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TINH NHANH :
A= ( -1/2)-(-3/5)+(-1/9)+1/131-(-2/7)+4/35-7/18
B=1/99-1/99.98-1/98.97-...-1/4.3-1/3.2-1/2.1
Bài 1 Rút gọn
a) \(1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-.....-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
Ta có : \(1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-......-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2013.2014}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=1-\left(1-\frac{1}{2014}\right)\)
\(=1-1+\frac{1}{2014}\)
\(=\frac{1}{2014}\)
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\(a,1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=1-\left(1-\frac{1}{2014}\right)\)
\(=1-1+\frac{1}{2014}\)
\(=\frac{1}{2014}\)
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Cho:
P=\(\frac{1}{2000.1999}-\frac{1}{1999.1998}-\frac{1}{1998.1997}-....................-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
Tính P+\(\frac{1997}{1999}\)
\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{1998.1999}\right)\)
\(=\frac{1}{2000.1999}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(=\frac{1}{2000.1999}-\left(1-\frac{1}{1999}\right)\)
\(=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow P+\frac{1997}{1999}=\frac{1}{1999.2000}-\frac{1998}{1999}+\frac{1997}{1999}\)
\(=\frac{-1}{2000}\)
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P= \(\frac{1}{2000.1999}\)- (\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\))
= \(\frac{1}{1999}-\frac{1}{2000}\)- (\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\))
= \(\frac{1}{1999}-\frac{1}{2000}\)- ( \(1-\frac{1}{1999}\))
= \(\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
= \(\frac{-1997}{1999}-\frac{1}{2000}\)
=) P + \(\frac{1997}{1999}\)= \(\frac{-1997}{1999}-\frac{1}{2000}+\frac{1997}{1999}=\frac{-1}{2000}\)
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Tính: P=-1+\(\dfrac{1}{2.1}+\dfrac{1}{3.2}+\dfrac{1}{4.3}+....+\dfrac{1}{2017.2016}+\dfrac{1}{2017}\)
Giúp mình với ae mai phải thi rùi
\(P=-1+\dfrac{1}{2.1}+\dfrac{1}{3.2}+..........+\dfrac{1}{2017.2016}+\dfrac{1}{2017}\)
\(=-1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{2016.2017}+\dfrac{1}{2017}\)
\(=-1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.........+\dfrac{1}{2016}-\dfrac{1}{2017}+\dfrac{1}{2017}\)
\(=-1+1-\dfrac{1}{2017}+\dfrac{1}{2017}\)
\(=0\)
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Bình luận (1)
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+\frac{1}{1998.1997}+\frac{1}{1997.1996}+...+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1} \)
Tính giá trị của biểu thức P.
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)
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Bình luận (3)
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+..+\frac{1}{3.2}+\frac{1}{2.1}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{1999}-\frac{1}{2000}\)
=\(1-\frac{1}{2000}\)
=\(\frac{1999}{2000}\)
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Bình luận (2)
$P = \dfrac1{2000 \cdot 1999} + \dfrac1{1999 \cdot 1998} + \ldots + \dfrac1{3 \cdot 2} + \dfrac1{2 \cdot 1} \\
= \dfrac1{1999} - \dfrac1{2000} + \dfrac1{1998} - \dfrac1{1999} + \ldots + \dfrac12 - \dfrac13 + \dfrac11 - \dfrac12
= - \dfrac1{2000} + \dfrac11 \\
= \dfrac{1999}{2000}$
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