\(\dfrac{2}{1009}\) \(\dfrac{6}{1009}\) \(\dfrac{12}{1009}\) \(\dfrac{20}{1009}\) \(......\)
Những câu hỏi liên quan
\(A=\left(\dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009}\right):\left(\dfrac{1}{23}+\dfrac{1}{7}-\dfrac{1}{1009}+\dfrac{1}{7}.\dfrac{1}{23}.\dfrac{1}{1009}\right)+1:\left(30.1009-160\right)\)
tìm giá trị biểu thức sau :
A=(\(\dfrac{1}{7}\)+\(\dfrac{1}{23}-\dfrac{1}{1009}\)):(\(\dfrac{1}{23}+\dfrac{1}{7}-\dfrac{1}{1009}+\dfrac{1}{7}.\dfrac{1}{23}.\dfrac{1}{1009}\))+1:(30.1009-160)
\(\dfrac{1}{9}-\dfrac{1}{1009}\)
\(A=\dfrac{2016^2+1^2}{2016\cdot1}+\dfrac{2015^2+2^2}{2015\cdot1}+\dfrac{2014^2+3^2}{2014\cdot3}+...+\dfrac{1009^2+1008^2}{1009\cdot1008}\)
và \(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}\)Tìm A/B
Giải phương trình: \(P=\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
ta có :
\(\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)
hay \(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\Leftrightarrow x-2010=0\)
hay x =2010
Vậy phương trình có nghiệm x = 2010
\(\dfrac{x-1009}{1001}+\dfrac{x+4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\Leftrightarrow\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\Leftrightarrow\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Rightarrow x=2010\)
Vậy....
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\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(x-2010=0\)
\(x=2010\)
Vậy x = 2010
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\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}=...+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)
\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+.....+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+........+\dfrac{x+2014}{1}\)\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+...+\left(\dfrac{x+1007}{1008}+1\right)=\left(\dfrac{x+1008}{1007}+1\right)+\left(\dfrac{x+1009}{1006}+1\right)+...+\left(\dfrac{x+2014}{1}+1\right)\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+1007}{1008}=\dfrac{x+2015}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+2015}{1008}-\dfrac{x+1008}{1007}-\dfrac{x+2015}{1006}-...-\dfrac{x+2015}{1}=0\)\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1\right)=0\)\(\Leftrightarrow x+2015=0\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1>0\right)\)\(\Leftrightarrow x=-2015\)
Vậy x=-2015
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\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\) Tìm n?
Ta có \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\)
\(\Leftrightarrow\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2n\left(2n+2\right)}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{n}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow2019n=1009\left(2n+2\right)\)
\(\Leftrightarrow2019n=2018n+2018\)
\(\Leftrightarrow n=2018\)
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so sánh \(\dfrac{1}{\sqrt{1.2017}}+\dfrac{1}{\sqrt{2.2016}}+...+\dfrac{1}{\sqrt{2017.1}}\) với \(\dfrac{2017}{1009}\)
Sắp xếp từ bé đến lớn các p/s sau : \(\dfrac{1005}{2002};\dfrac{1007}{2006};\dfrac{1009}{2010};\dfrac{1011}{2014}\)