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\(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(P=\frac{1}{1999.2000}-\frac{1}{1998.1999}-...-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=\frac{1}{1999}-\frac{1}{2000}-\frac{1}{1998}+\frac{1}{1999}-\frac{1}{1997}+\frac{1}{1998}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(P=\frac{2}{1999}-\frac{1}{2000}-1\)
\(P+\frac{1997}{1999}=\frac{2}{1999}+\frac{1997}{1999}-\frac{1}{2000}-1=1-1-\frac{1}{2000}=-\frac{1}{2000}\)
Số hạng đầu tiên không theo quy luật hả (+) hày (-) đề thế nào làm vậy:
\(P=\frac{1}{2000.1998}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{1998.1999}\right)=\frac{1}{1999.2000}-Q\)
Tổng quát ta có \(\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}\) với dãy trên ta luôn có b-a=1
\(Q=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-.....-\frac{1}{1999}\)
\(Q=1-\frac{1}{1999}\Rightarrow P=\frac{1}{1999.2000}-1+\frac{1}{1999}=\frac{1-1999.2000+2000}{1999.2000}=\frac{1-1998.2000}{1999.2000}\)
\(P+\frac{1997}{1998}=\frac{1997}{1998}+\frac{1-1998.2000}{1999.2000}\) xem lại đề
Hôm kia giải thi chơi được 260, làm được bài này luôn. Hôm sau, làm lại chả biết làm.
\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{1998.1999}\right)\)
\(=\frac{1}{2000.1999}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(=\frac{1}{2000.1999}-\left(1-\frac{1}{1999}\right)\)
\(=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow P+\frac{1997}{1999}=\frac{1}{1999.2000}-\frac{1998}{1999}+\frac{1997}{1999}\)
\(=\frac{-1}{2000}\)
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 + \(\frac{1997}{1999}\)= \(\frac{-1997}{1999}-\frac{1}{2000}+\frac{1997}{1999}=\frac{-1}{2000}\)
Ta có:
\(P=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow P=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow P=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow P=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow P=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow P=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow P=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow P=\frac{-1997}{1999}-\frac{1}{2000}\)
\(\Rightarrow P+\frac{1997}{1999}=\frac{-1997}{1999}-\frac{1}{2000}+\frac{1}{1997}\)
\(\Rightarrow P+\frac{1997}{1999}=\frac{-1}{2000}\)
Vậy....
khó quá mik lm hk ra ai giỏi toán giúp mik với!
b,
\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)
3) 2x3-1=15 <=> x3=16/2=8=23 => x=2
\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{x+16+y-25+z+9}{9+16+25}=\frac{x+y+z}{50}\)
=> \(\frac{x+16}{9}=\frac{x+y+z}{50}\)=> x+y+z=\(\frac{50\left(x+16\right)}{9}\)=\(\frac{50\left(2+16\right)}{9}=\frac{50.18}{9}=50.2=100\)
Vậy x+y+z=100
C= \(\dfrac{1}{100}-\)(\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{98.99}\)+\(\dfrac{1}{99.100}\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
=\(\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
= \(\dfrac{1}{100}-\dfrac{99}{100}\)
=\(\dfrac{-98}{100}=-\dfrac{49}{50}\)
Ta có:
\(=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+......+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)
sau khi giản ước ta được như sau:
=\(\dfrac{1}{100}-1\)=\(\dfrac{-99}{100}\)
\(F=-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\)
\(\Rightarrow-F=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}\)\(\Rightarrow F=\dfrac{-2015}{2016}\)
Giải:
\(F=\dfrac{-1}{2016.2015}-\dfrac{1}{2015.2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(\Leftrightarrow F=-\left(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+\dfrac{1}{2013.2012}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}+\dfrac{1}{2013.2014}+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2016}\right)\)
\(\Leftrightarrow F=-\dfrac{2015}{2016}\)
Vậy ...
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)
\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+..+\frac{1}{3.2}+\frac{1}{2.1}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{1999}-\frac{1}{2000}\)
=\(1-\frac{1}{2000}\)
=\(\frac{1999}{2000}\)
-1/2000
\(P=\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(\Rightarrow P=\dfrac{1}{1999.2000}-\dfrac{1}{1998.1999}-\dfrac{1}{1997.1998}-\dfrac{1}{2.3}-\dfrac{1}{1.2}\)
\(\Rightarrow P=\dfrac{1}{1999}-\dfrac{1}{2000}-\dfrac{1}{1998}+\dfrac{1}{1999}-\dfrac{1}{1997}+\dfrac{1}{1998}-...-1+\dfrac{1}{2}\)
\(\Rightarrow P=\dfrac{2}{1999}-\dfrac{1}{2000}-1\)
\(\Rightarrow P+\dfrac{1997}{1999}=\dfrac{2}{1999}+\dfrac{1997}{1999}-\dfrac{1}{2000}-1\)
\(\Rightarrow P+\dfrac{1997}{1999}=1-1-\dfrac{1}{2000}=\dfrac{-1}{1200}\)
Vậy \(P+\dfrac{1997}{1999}=\dfrac{-1}{2000}\)